NCERT Solutions for Class 12 Physics Chapter 9: Ray Optics and Optical Instruments

Concept and context. A concave mirror is a spherical mirror whose reflecting surface faces the centre of curvature; it converges a parallel beam to its principal focus F, which lies midway between the pole P and the centre of curvature C: f = R/2. The position, nature and size of the image follow from the mirror equation 1/v + 1/u = 1/f and linear magnification m = h′/h = −v/u, using the Cartesian sign convention (distances measured from the pole; those against the direction of the incident light — i.e. in front of the mirror — are negative; heights above the axis positive). For a concave mirror f is negative. An image that can be obtained on a screen is a real image (reflected rays actually meet); a virtual image (rays only appear to meet) cannot be taken on a screen.

Given: h = 2.5 cm; u = −27 cm; R = −36 cm (concave) ⇒ f = R/2 = −18 cm.

Step 1 — Position of the image (and of the screen)

1. Mirror equation: 1/v = 1/f − 1/u = 1/(−18) − 1/(−27) = −1/18 + 1/27.
2. Common denominator 54: 1/v = (−3 + 2)/54 = −1/54.
3. v = −54 cm. The negative sign shows the image is in front of the mirror (on the same side as the object), i.e. it is real. The screen must therefore be placed 54 cm from the mirror, on the same side as the candle.

Step 2 — Size and nature of the image

1. Magnification: m = −v/u = −(−54)/(−27) = −2.
2. Image height: h′ = m h = −2 × 2.5 = −5.0 cm. The negative sign means the image is inverted.
3. Nature: real, inverted and magnified (twice the size of the candle, 5.0 cm tall). The object lies between F (18 cm) and C (36 cm), and the image forms beyond C (54 cm > 36 cm).

Step 3 — Effect of moving the candle closer to the mirror

• As the object moves from 27 cm towards the focus (18 cm), 1/v = 1/f − 1/u shows that |v| increases: e.g. u = −24 cm ⇒ v = −72 cm; u = −20 cm ⇒ v = −180 cm. So the screen has to be moved away from the mirror (the image also becomes larger, m = −v/u = −3 at u = −24 cm).
• When the candle reaches the focus (u = −18 cm), the image goes to infinity; if it is moved closer than F (between F and P) the image becomes virtual, erect and magnified behind the mirror and can no longer be obtained on a screen.

• Check with the ray diagram: a ray parallel to the axis passes through F after reflection, a ray through C retraces its path; they meet at the tip of the image 54 cm in front of the mirror, below the axis — inverted.
• Uses: a real, magnified image with the object between F and C is the principle of a projector and of the objective of a reflecting telescope.

Conclusion: the screen must be placed 54 cm from the mirror; the image is real, inverted and magnified (5.0 cm tall, m = −2); if the candle is moved closer to the mirror (towards F), the image moves farther away and grows larger, so the screen has to be moved away from the mirror.

Concept and context. A convex mirror has its reflecting surface bulging towards the object; it always diverges a beam, so a parallel beam appears to come from the principal focus F behind the mirror. By the Cartesian sign convention the focal length of a convex mirror is positive (f = +R/2), the object distance u is negative (object in front). Using the mirror equation 1/v + 1/u = 1/f with a positive f always gives a positive v: the image is virtual (formed behind the mirror), erect and diminished. These properties make convex mirrors useful as rear-view mirrors and in shop security mirrors (wide field of view).

Given: h = 4.5 cm; u = −12 cm; f = +15 cm.

1. Image position: 1/v = 1/f − 1/u = 1/15 + 1/12 = (4 + 5)/60 = 9/60 = 3/20.
2. v = 20/3 = +6.67 cm — positive, so the image is behind the mirror, 6.7 cm from the pole (between P and F, since 6.7 < 15 cm), and it is virtual.
3. Magnification: m = −v/u = −(6.67)/(−12) = +0.56.
4. Size of the image: h′ = m h = 0.56 × 4.5 = 2.5 cm, erect (positive) and diminished.

v = +6.7 cm (behind the mirror) ; m = +0.56 ; image size 2.5 cm, virtual, erect

As the needle is moved farther from the mirror

• The image moves towards the focus (v → +15 cm as u → −∞), i.e. it shifts away from the pole but never goes beyond F: for u = −30 cm, v = +10 cm; u = −60 cm, v = +12 cm; u = −150 cm, v = +13.6 cm.
• The image remains virtual and erect and becomes smaller (m = −v/u decreases: 0.56 → 0.33 → 0.2 → 0.09), so the needle appears progressively tinier while being squeezed towards the focus.

Conclusion: the image is virtual, erect and diminished, formed 6.7 cm behind the mirror with magnification +0.56 (size 2.5 cm); as the needle is moved farther away the image moves towards the focus and becomes smaller.

Concept and context. When an object under a transparent liquid is viewed from above (near the normal), refraction at the liquid surface makes it appear raised: the rays from the object bend away from the normal on emerging into air, so the eye sees an image at a smaller depth than the real depth. For near-normal viewing, refractive index n = real depth / apparent depth (derived from Snell’s law with small angles: n = sin i/sin r ≈ tan i/tan r = (x/AP’)/(x/AP) …). A travelling microscope measures apparent depth by focusing first on the surface marking and then on the image of the object (NCERT Fig. 9.10).

Given: real depth d = 12.5 cm; apparent depth in water = 9.4 cm; liquid index n′ = 1.63 filled to the same height 12.5 cm.

(i) Refractive index of water

n = real depth/apparent depth = 12.5/9.4 = 1.33

(ii) Shift of the microscope for the liquid of index 1.63

1. Apparent depth in the new liquid: d′ = real depth/n′ = 12.5/1.63 = 7.67 cm.
2. Change in apparent depth: 9.4 − 7.67 = 1.73 cm — the image of the needle is now nearer to the surface (shallower) because the liquid is optically denser (greater n).
3. Hence the microscope has to be moved downwards (towards the liquid) by about 1.7 cm to focus on the needle again.

• Derivation of n = d/d′: for a ray from the object O at depth d leaving the surface at a small angle, Snell’s law n sin i = sin r gives n ≈ tan r/tan i = (x/d′)/(x/d) = d/d′ ⇒ the image appears at d′ = d/n.
• Everyday effects: a swimming pool appears shallower than it is; a straw appears bent at the water surface; fish appear closer to the surface.

Conclusion: the refractive index of water is 1.33; for the liquid of index 1.63 the apparent depth is 7.67 cm, so the microscope has to be moved downward by ≈ 1.7 cm to focus on the needle again.

Concept and context. Refraction at the interface of two media obeys Snell’s law: n₁ sin i = n₂ sin r (the incident ray, the refracted ray and the normal are coplanar), where n₁ and n₂ are the refractive indices (absolute, relative to vacuum) of the first and second media. From two refractions with respect to a common medium (air) we can determine the refractive indices of glass and water relative to air and hence the relative index of glass with respect to water — then apply Snell’s law at the water–glass interface. This is also an application of the relation n₃₁ = n₃₂ × n₂₁ (chain rule) for refractive indices.

Data read from Fig. 9.27: (a) air → glass: i = 60°, r = 35°; (b) air → water: i = 60°, r = 47°; (c) water → glass: i = 45°, r = ?

1. Refractive index of glass (relative to air): n_g = sin 60°/sin 35° = 0.8660/0.5736 = 1.51.
2. Refractive index of water (relative to air), from the figure: n_w = sin 60°/sin 47° = 0.8660/0.7314 = 1.18.
3. Snell’s law at the water–glass interface: n_w sin 45° = n_g sin r ⇒ sin r = (n_w/n_g) sin 45° = (1.184/1.510) × 0.7071 = 0.7842 × 0.7071 = 0.5545.
4. Angle of refraction in glass: r = sin⁻¹(0.5545) ≈ 33.7° (≈ 34°).

r ≈ 34° (sin r = n_w sin 45° / n_g = 0.55)

• Relative index of glass w.r.t. water: n_gw = n_g/n_w = 1.51/1.18 = 1.28; so sin r = sin 45°/1.28 = 0.55 ✓. The ray bends towards the normal (45° → 34°) in going from the rarer (water) to the denser (glass) medium, as expected.
• Remark on the data: the figure gives r = 47° for the air–water refraction, which corresponds to n_w = 1.18; the actual refractive index of water is 1.33 (which would give 40.6° in (b)). Using the actual n_w = 1.33 with n_g = 1.51 would give sin r = 1.33 × 0.7071/1.51 = 0.623, r ≈ 38.5°. The answer ≈ 34° follows from the data as printed in the figure.
• Method (general): n₁ sin i₁ = n₂ sin i₂ for any pair: sin i/sin r = n₂/n₁ — the common medium (air) is used to find each index relative to vacuum (air ≈ vacuum).

Conclusion: the angle of refraction in glass for a ray incident at 45° in water is about 34° (sin r = n_w sin 45°/n_g with n_g = 1.51 and n_w = 1.18 from the data of the figure).

Concept and context. Light from a point source under water travels in all directions; at the water–air surface a ray emerges only if its angle of incidence is less than the critical angle i_c. For a ray going from a denser medium (water, n = 1.33) to a rarer medium (air), the angle of refraction becomes 90° at the critical angle: sin i_c = 1/n. Rays striking the surface at angles greater than i_c undergo total internal reflection (TIR) and stay in the water. The rays that can emerge therefore fill a cone of semi-vertical angle i_c with its apex at the bulb; this cone cuts the surface in a circle of radius r = d tan i_c (d = depth).

Given: depth d = 80 cm = 0.80 m; n = 1.33.

1. Critical angle: sin i_c = 1/n = 1/1.33 = 0.7519 ⇒ i_c = sin⁻¹(0.7519) = 48.75° (≈ 48.8°).
2. tan i_c: cos i_c = √(1 − 0.5653) = 0.6593; tan i_c = sin i_c/cos i_c = 0.7519/0.6593 = 1.140. (Equivalently, tan i_c = 1/√(n² − 1) = 1/√(1.769 − 1) = 1/0.877 = 1.140.)
3. Radius of the illuminated circle: r = d tan i_c = 0.80 × 1.140 = 0.912 m (≈ 91 cm).
4. Area: A = π r² = 3.14 × (0.912)² = 3.14 × 0.832 = 2.6 m².

A = π d² / (n² − 1) = 2.6 m²

• Why a circle? The bulb is a point source and the water surface is flat, so by symmetry the emergent light fills a circular patch centred directly above the bulb.
• Outside the circle the light is totally internally reflected back into the water — which is why a diver looking up sees the sky only through a circular “window” of semi-angle 48.8° (97.5° field of view), the rest of the surface acting as a mirror.
• Depth dependence: A ∝ d² — for 40 cm depth the area would be 0.65 m².

Conclusion: light from the bulb can emerge through a circular area of the water surface of radius ≈ 0.91 m, i.e. an area of about 2.6 m²; elsewhere it is totally internally reflected.

Concept and context. A ray passing through a prism of refracting angle A is deviated by δ = i + e − A, where i and e are the angles of incidence and emergence (and r₁ + r₂ = A). The deviation varies with the angle of incidence; it is a minimum, δ_m, when the ray passes symmetrically (i = e, r₁ = r₂ = A/2). The refractive index of the prism material relative to the surrounding medium is then n₂₁ = sin[(A + δ_m)/2] / sin(A/2) (NCERT Fig. 9.21–9.22). When the prism is immersed in a denser medium such as water, the relative refractive index n₂₁ = n_glass/n_water decreases, the bending is weaker and the angle of minimum deviation is smaller.

Given: A = 60°; δ_m = 40°; n_water = 1.33.

(i) Refractive index of the glass

1. (A + δ_m)/2 = (60° + 40°)/2 = 50°; A/2 = 30°.
2. n = sin 50°/sin 30° = 0.7660/0.5000 = 1.53.

(ii) Minimum deviation of the prism in water

1. Relative refractive index of glass w.r.t. water: n_gw = n_g/n_w = 1.532/1.33 = 1.152.
2. Minimum-deviation relation: n_gw = sin[(A + δ_m′)/2]/sin(A/2) ⇒ sin[(60° + δ_m′)/2] = 1.152 × sin 30° = 1.152 × 0.5 = 0.576.
3. (60° + δ_m′)/2 = sin⁻¹(0.576) = 35.2° ⇒ 60° + δ_m′ = 70.4° ⇒ δ_m′ ≈ 10.4° (about 10°).

n = 1.53 ; δ_m (in water) ≈ 10°

• Interpretation: the deviation drops from 40° to about 10° because the refractive index contrast between glass and water (1.15) is much smaller than that between glass and air (1.53). In a medium of the same refractive index as the glass the prism would produce no deviation (like the vanishing glass lens in Example 9.6).
• Accuracy: the glass index 1.532 was used without rounding; using 1.53 gives δ_m′ ≈ 10.3°.
• Thin-prism check: δ ≈ (n − 1)A gives 32° for glass in air — a rough estimate since A = 60° is not “thin”.

Conclusion: the refractive index of the glass is 1.53; when the prism is immersed in water the angle of minimum deviation reduces to about 10°.

Concept and context. The focal length of a thin lens is governed by the lens maker’s formula: 1/f = (n − 1)(1/R₁ − 1/R₂), where n is the refractive index of the lens material relative to the surrounding medium (air here), and R₁, R₂ are the radii of curvature of the first and second surfaces, with the Cartesian sign convention (distances measured in the direction of the incident light positive). For a double-convex (equiconvex) lens the first surface has its centre of curvature on the image side (R₁ = +R) and the second on the object side (R₂ = −R), so 1/R₁ − 1/R₂ = 2/R.

Given: n = 1.55; f = +20 cm; |R₁| = |R₂| = R.

1. Lens maker’s formula: 1/f = (n − 1)(1/R₁ − 1/R₂) = (n − 1)(1/R + 1/R) = 2(n − 1)/R.
2. Solve for R: R = 2(n − 1) f = 2 × (1.55 − 1) × 20 = 2 × 0.55 × 20.
3. R = 22 cm.

R = 2(n − 1)f = 22 cm

• Check: 1/f = 0.55 × (2/22) = 0.05 cm⁻¹ ⇒ f = 20 cm ✓.
• Power: P = 1/f = 1/0.20 m = +5 D.
• Remark: a glass of n = 1.5 would need R = 20 cm for the same f (R = f when n = 1.5); the larger the refractive index, the smaller the radius required.

Conclusion: each face of the double-convex lens must have a radius of curvature R = 22 cm.

Concept and context. A converging beam heading to a point P acts, for a lens placed in its path, as a virtual object located at P. By the Cartesian sign convention, distances are measured from the optical centre in the direction of the incident light as positive; since P lies beyond the lens (in the direction of travel of the light), the virtual object distance is positive: u = +12 cm. The thin-lens formula 1/v − 1/u = 1/f is then applied with f positive for a convex lens and negative for a concave lens. A positive v means the beam converges to a real image point beyond the lens.

Given: the lens is 12 cm from P: virtual object at u = +12 cm.

(a) Convex lens, f = +20 cm

1. 1/v = 1/f + 1/u = 1/20 + 1/12 = (3 + 5)/60 = 8/60 = 2/15.
2. v = 15/2 = +7.5 cm. The beam, now more strongly converging, meets at a point 7.5 cm from the lens on the side of P (4.5 cm before the original convergence point P).

(b) Concave lens, f = −16 cm

1. 1/v = 1/f + 1/u = −1/16 + 1/12 = (−3 + 4)/48 = 1/48.
2. v = +48 cm. The diverging action of the lens reduces the convergence of the beam: it now converges at a point 48 cm from the lens, on the same side as P (36 cm beyond P).

• Interpretation using vergence (power): the incident beam has vergence +1/0.12 = +8.33 D; the convex lens adds +5 D → 13.3 D ⇒ v = 7.5 cm; the concave lens adds −6.25 D → +2.08 D ⇒ v = 48 cm.
• Why positive u? A real object lies in front of the lens (u negative). Here the light has not yet converged when it meets the lens, so the “object” P lies beyond the lens: u is positive — a virtual object.

Conclusion: (a) with the convex lens the beam converges at 7.5 cm from the lens; (b) with the concave lens it converges at 48 cm from the lens (both on the side of the original point P).

Concept and context. A concave (diverging) lens is thinner at the centre than at the edges; it diverges a parallel beam so that the rays appear to come from its principal focus on the side of the incident light. By the Cartesian sign convention its focal length is negative (f = −21 cm) and a real object is at a negative distance (u = −14 cm). The thin-lens formula is 1/v − 1/u = 1/f and the linear magnification is m = h′/h = v/u. A concave lens always produces a virtual, erect, diminished image located between the optical centre and the first principal focus, wherever the (real) object is placed.

Given: h = 3.0 cm; u = −14 cm; f = −21 cm.

1. Image position: 1/v = 1/f + 1/u = −1/21 − 1/14 = −(2 + 3)/42 = −5/42 ⇒ v = −42/5 = −8.4 cm.
2. The negative sign means the image is on the same side as the object (the side from which the light comes), 8.4 cm from the lens — i.e. virtual.
3. Magnification: m = v/u = (−8.4)/(−14) = +0.6 ⇒ image height h′ = 0.6 × 3.0 = 1.8 cm.
4. Nature: virtual, erect (m positive) and diminished (1.8 cm < 3.0 cm), located 8.4 cm from the lens, between the lens and its focus (21 cm).

v = −8.4 cm (virtual, same side as object) ; m = +0.6 ; image size 1.8 cm, erect

Effect of moving the object farther from the lens

• As |u| increases the image moves towards the focus (v → −21 cm as u → −∞) — it remains virtual and erect, and its size decreases further (m → 0).
• Examples: u = −30 cm ⇒ 1/v = −1/21 − 1/30 = −17/210 ⇒ v = −12.4 cm, m = +0.41; u = −100 cm ⇒ v = −17.4 cm, m = +0.17.

• Consequences: a concave lens can never give a real image of a real object; this is why it is used in spectacles for myopia (short-sightedness) and in combinations (e.g. Galilean telescopes, zoom lenses) to produce a smaller virtual image.

Conclusion: the image is virtual, erect and diminished, 8.4 cm from the lens on the same side as the object, 1.8 cm tall (m = +0.6); as the object is moved farther away the image moves closer to the focus and becomes smaller.

Concept and context. When two thin lenses of focal lengths f₁ and f₂ are placed in contact (thickness ignored) the combination behaves like a single thin lens whose focal length F satisfies 1/F = 1/f₁ + 1/f₂ — equivalently the powers add: P = P₁ + P₂ (P = 1/f, in dioptres), and the total magnification is the product m = m₁ m₂ (NCERT Section 9.5.4, Fig. 9.19). The image formed by the first lens is the object for the second. A convex lens has positive f and a concave lens negative f; the sign of the net power decides whether the system is converging (P > 0) or diverging (P < 0).

Given: f₁ = +30 cm (convex); f₂ = −20 cm (concave).

1. Combination formula: 1/F = 1/f₁ + 1/f₂ = 1/30 − 1/20.
2. Common denominator 60: 1/F = (2 − 3)/60 = −1/60.
3. F = −60 cm.

Nature of the system

• The focal length is negative, so the combination is a diverging (concave) lens of focal length 60 cm: the concave lens (power −5 D) is stronger than the convex one (+3.33 D).
• Power: P = P₁ + P₂ = +3.33 D − 5.0 D = −1.67 D (= 1/(−0.60 m)).
• Effect on a parallel beam: it emerges diverging, appearing to come from a point 60 cm from the lenses on the incident side.

F = −60 cm ; P = −1.67 D (diverging)

• Condition for a converging system: |f_convex| < |f_concave| (the convex lens must be stronger); if |f₁| = |f₂| the combination has zero power (a plane glass plate, F = ∞).
• Application: achromatic doublets (a convex and a concave lens of different glasses in contact) cancel chromatic aberration while keeping a net converging power.

Conclusion: the combination has a focal length of −60 cm and is therefore a diverging lens.

Concept and context. In a compound microscope (NCERT Fig. 9.24) the objective (short focal length f_o) forms a real, inverted, magnified image A′B′ of the small object placed just beyond its focus; this image acts as the object for the eyepiece (f_e), which behaves as a simple magnifier and forms a virtual, magnified final image. The total magnifying power is the product M = m_o × m_e, where m_o = v_o/u_o is the linear magnification of the objective and m_e is the angular magnification of the eyepiece: m_e = 1 + D/f_e if the final image is at the near point D = 25 cm, and m_e = D/f_e if the final image is at infinity (relaxed eye). The distance between the lenses = v_o + |u_e|.

Given: f_o = 2.0 cm; f_e = 6.25 cm; separation L = 15 cm; D = 25 cm.

(a) Final image at the least distance of distinct vision (25 cm)

1. Eyepiece: final (virtual) image at v_e = −25 cm: 1/u_e = 1/v_e − 1/f_e = −1/25 − 1/6.25 = −0.04 − 0.16 = −0.20 ⇒ u_e = −5.0 cm. So the image formed by the objective must be 5 cm in front of the eyepiece.
2. Objective: the image distance v_o = separation − |u_e| = 15 − 5 = 10 cm.
3. Object distance for the objective: 1/u_o = 1/v_o − 1/f_o = 1/10 − 1/2 = −0.4 ⇒ u_o = −2.5 cm. The object must be placed 2.5 cm from the objective.
4. Magnifying power: m_o = v_o/u_o = 10/(−2.5) = −4; m_e = 1 + D/f_e = 1 + 25/6.25 = 5 ⇒ M = m_o m_e = −20 (magnitude 20; negative sign: inverted final image).

(b) Final image at infinity

1. Eyepiece: for the image at infinity the intermediate image must be at the focus of the eyepiece: |u_e| = f_e = 6.25 cm.
2. Objective image distance: v_o = 15 − 6.25 = 8.75 cm.
3. Object distance: 1/u_o = 1/8.75 − 1/2 = 0.11429 − 0.5 = −0.38571 ⇒ u_o = −2.59 cm (object 2.59 cm from the objective).
4. Magnifying power: m_o = v_o/u_o = 8.75/(−2.59) = −3.375; m_e = D/f_e = 25/6.25 = 4 ⇒ M = −3.375 × 4 = −13.5 (magnitude 13.5).

• Observation: the magnifying power is greater when the final image is at the near point (20) than at infinity (13.5) — but viewing at infinity is more comfortable (relaxed eye, no strain).
• Quick approximate formula: M ≈ (L/f_o)(D/f_e) with L the distance between the second focus of the objective and the first focus of the eyepiece: (15 − 2 − 6.25)/2 × 4 = 13.5 ✓ for case (b).
• The object distance is only slightly greater than f_o (2.0 cm), as required to obtain a highly magnified real image.

Conclusion: (a) the object should be 2.5 cm from the objective with M = −20; (b) the object should be 2.59 cm from the objective with M = −13.5.

Concept and context. For a person with a normal eye the near point is D = 25 cm; “sharp focus” with the eye relaxed to its near point means the final (virtual) image is at 25 cm from the eyepiece. The separation between the lenses equals v_o + |u_e|, where v_o is the image distance for the objective and u_e is the object distance for the eyepiece (the intermediate image). The magnifying power is M = m_o × m_e with m_o = v_o/u_o and m_e = 1 + D/f_e (image at the near point).

Given: f_o = 8.0 mm = 0.80 cm; f_e = 2.5 cm; u_o = −9.0 mm = −0.90 cm; D = 25 cm; final image at 25 cm.

1. Objective: 1/v_o = 1/f_o + 1/u_o = 1/0.80 − 1/0.90 = 1.25 − 1.1111 = 0.1389 ⇒ v_o = 7.2 cm.
2. Eyepiece: final image at v_e = −25 cm: 1/u_e = 1/v_e − 1/f_e = −1/25 − 1/2.5 = −0.04 − 0.40 = −0.44 ⇒ u_e = −2.27 cm.
3. Separation of the lenses: L = v_o + |u_e| = 7.20 + 2.27 = 9.47 cm.
4. Magnifying power: m_o = v_o/u_o = 7.2/(−0.9) = −8; m_e = 1 + D/f_e = 1 + 25/2.5 = 11 ⇒ M = −8 × 11 = −88.

separation = 9.47 cm ; M = −88

• Sign: the negative sign of M indicates that the final image is inverted relative to the object (as in all compound microscopes).
• Check: approximate formula M ≈ (L/f_o)(D/f_e) with L the tube length (v_o − f_o = 6.4 cm): (6.4/0.8)(25/2.5) = 80 — close to the exact value of 88.

Conclusion: the lenses are separated by 9.47 cm and the magnifying power of the microscope is −88 (magnitude 88).

Concept and context. In an astronomical (refracting) telescope (NCERT Fig. 9.25) the objective of long focal length f_o forms a real, inverted image of a distant object at its focal plane; the eyepiece of short focal length f_e is positioned so that this image lies at (or just within) its focus and produces the final virtual image. The magnifying power (angular magnification) is the ratio of the angle β subtended at the eye by the final image to the angle α subtended at the eye by the object: M = β/α = f_o/f_e (for the telescope in normal adjustment, final image at infinity), and the length of the telescope (separation of the lenses) is L = f_o + f_e.

Given: f_o = 144 cm; f_e = 6.0 cm.

1. Magnifying power: M = f_o/f_e = 144/6.0 = 24.
2. Separation between the lenses (normal adjustment): L = f_o + f_e = 144 + 6.0 = 150 cm.

M = f_o/f_e = 24 ; L = f_o + f_e = 150 cm

• Meaning of M = 24: a distant object appears 24 times larger in angular size (it subtends an angle 24 times greater at the eye than it does for the unaided eye).
• Design principle: to obtain a high magnifying power use an objective of large f_o (and large aperture to collect more light and improve resolving power) with an eyepiece of small f_e.
• Final image: at infinity; virtual and inverted.

Conclusion: the magnifying power of the telescope is 24 and the objective and the eyepiece are separated by 150 cm.

Concept and context. For a refracting telescope in normal adjustment the angular magnification is M = f_o/f_e. The objective forms a real, inverted image of the distant object in its focal plane. A distant object of diameter d at distance r subtends an angle α = d/r at the objective; the image formed by the objective has a linear size d′ = f_o α = f_o (d/r), since the image of a point at infinity in the direction α lies at height f_o α from the axis in the focal plane. The Moon, though large (3.48 × 10⁶ m), is so far away (3.8 × 10⁸ m) that it subtends only about half a degree.

Given: f_o = 15 m; f_e = 1.0 cm = 0.010 m; Moon’s diameter d = 3.48 × 10⁶ m; orbit radius r = 3.8 × 10⁸ m.

(a) Angular magnification

M = f_o/f_e = 15 m / 0.010 m = 1500

(b) Diameter of the image of the Moon formed by the objective

1. Angle subtended by the Moon: α = d/r = (3.48 × 10⁶)/(3.8 × 10⁸) = 9.16 × 10⁻³ rad (≈ 0.52° or 31′).
2. Diameter of the image in the focal plane: d′ = f_o α = 15 × 9.16 × 10⁻³ = 0.137 m.
3. d′ ≈ 13.7 cm.

• Equivalent form (lens formula): m = v/u = f_o/r (object at a great distance r, v ≈ f_o) ⇒ d′ = d f_o/r = 3.48 × 10⁶ × 15/3.8 × 10⁸ = 0.137 m ✓.
• Use of the objective image: the eyepiece then magnifies the angular size of this 13.7 cm real image by 1500 (the Moon’s angular diameter of ≈ 0.52° would become about 780° if the whole disc could be seen). In practice the field of view of the eyepiece is limited, so only a small portion of the Moon’s surface is seen at this magnification.
• Why such large telescopes? A large f_o gives a large image of the Moon (13.7 cm) and a large aperture collects more light.

Conclusion: (a) the angular magnification is 1500; (b) the image of the Moon formed by the objective has a diameter of about 13.7 cm.

Concept and context. This exercise shows how the properties of images that one gets from explicit ray diagrams follow algebraically from the mirror equation 1/v + 1/u = 1/f and the magnification m = −v/u, with the Cartesian sign convention: distances are measured from the pole P; those along the direction of the incident light are positive, so the object distance u is negative for a real object, the focal length f is negative for a concave mirror and positive for a convex mirror; a negative v (in front of the mirror) means a real image and a positive v (behind the mirror) a virtual image. To avoid sign confusion write f = −F (F > 0) for the concave mirror, u = −x (x > 0) for the object and work with magnitudes F, x.

(a) Object between f and 2f of a concave mirror ⇒ real image beyond 2f

1. Let the object be at distance x from the pole with F < x < 2F. The mirror equation 1/v = 1/f − 1/u becomes 1/v = −1/F + 1/x = (F − x)/(F x) (using f = −F, u = −x).
2. Since x > F, F − x < 0 ⇒ v is negative: v = −Fx/(x − F). A negative v means the image is in front of the mirror ⇒ real.
3. Position: |v| = Fx/(x − F). The condition |v| > 2F is Fx > 2F(x − F) ⇔ x > 2x − 2F ⇔ x < 2F, which is true by assumption. Hence |v| > 2F: the image is real and beyond 2F (the centre of curvature). (Numerical illustration: F = 10 cm, x = 15 cm ⇒ v = −30 cm.)
4. Magnification m = −v/u = −(−Fx/(x − F))/(−x) = −F/(x − F), magnitude > 1 (since x − F < F): enlarged and negative (inverted).

(b) A convex mirror always produces a virtual image

1. For a convex mirror f = +f (positive) and u = −x (x > 0, a real object in front): 1/v = 1/f − 1/u = 1/f + 1/x.
2. Both terms are positive, so 1/v > 0 for every x > 0 ⇒ v is always positive — the image lies behind the mirror and is virtual, irrespective of the object’s position.

(c) The virtual image of a convex mirror is diminished and lies between the focus and the pole

1. From (b): v = fx/(f + x), which is positive. Since f + x > x, v < f; since f + x > f, v < x (i.e. v < |u|). So 0 < v < f: the image lies between the pole and the focus.
2. Magnification: m = −v/u = v/x = f/(f + x) < 1, and positive. Hence the image is diminished and erect for every position of the object; as x → ∞, v → f and m → 0; as x → 0, v → 0 and m → 1.

(d) An object between the pole and the focus of a concave mirror gives a virtual enlarged image

1. Take the object at x from the pole with 0 < x < F: 1/v = −1/F + 1/x = (F − x)/(F x) > 0 since F > x ⇒ v = Fx/(F − x) is positive: image behind the mirror ⇒ virtual.
2. Magnification: m = −v/u = v/x = F/(F − x). Since 0 < F − x < F, m > 1 and positive: the image is magnified and erect. (Examples: F = 15 cm, x = 5 cm ⇒ v = +7.5 cm, m = +1.5; NCERT Example 9.3(ii): F = 7.5 cm, x = 5 cm ⇒ v = +15 cm, m = +3.)

• Everyday uses: (a) projector/telescope mirror; (b),(c) convex rear-view mirrors with a wide field of view; (d) shaving or make-up mirrors and the dentist’s mirror give an enlarged erect image.

Conclusion: from 1/v = 1/f − 1/u: (a) F < x < 2F gives v negative with |v| > 2F (real, beyond 2F); (b) a convex mirror gives 1/v > 0 always (virtual); (c) its image has v = fx/(f + x) < f and m = f/(f + x) < 1; (d) for x < F, v > 0 and m = F/(F − x) > 1 (virtual and enlarged).

Concept and context. A plane parallel slab of thickness t and refractive index n makes an object seen through it (near normal incidence) appear raised because of refraction: the part of the path inside the glass is “shortened” to t/n. The apparent shift (the distance by which the object appears to be raised towards the observer) is Δ = t(1 − 1/n). The same result arises from the apparent-depth relation (apparent thickness = t/n, so the shift is t − t/n). The air path outside the slab is unaffected (the refractive index of air ≈ 1).

Given: t = 15 cm; n = 1.5; viewing distance 50 cm (irrelevant to the shift).

1. Apparent thickness of the slab: t/n = 15/1.5 = 10 cm.
2. Shift (rise) of the pin: Δ = t − t/n = 15 − 10 = 5 cm, i.e. Δ = t(1 − 1/n) = 15 × (1 − 2/3) = 5 cm.
3. The pin therefore appears 5 cm closer to the observer — at a distance of 45 cm instead of 50 cm.

Does the answer depend on the location of the slab?

• No. The shift Δ = t(1 − 1/n) depends only on the thickness t of the slab and its refractive index n — not on where the slab is placed between the eye and the pin (provided it is parallel to the table and the viewing is near-normal), because the air gaps above and below the slab contribute their real distances regardless of the slab position and the slab only replaces a length t by an apparent length t/n. If the slab is placed anywhere in between, the pin still appears raised by 5 cm.

Δ = t(1 − 1/n) = 5 cm, independent of the slab position

Conclusion: the pin appears raised by 5 cm, and this shift does not depend on the location of the slab.

Concept and context. A light pipe (optical fibre) guides light by repeated total internal reflection (TIR) at the interface between the glass core (index n₁ = 1.68) and the outer covering or cladding (index n₂ = 1.44 < n₁). TIR at the core–cladding surface requires the angle of incidence there to exceed the critical angle i_c given by sin i_c = n₂/n₁. The ray enters through the end face at an angle of incidence i with the axis and refracts to an angle r; because the core–cladding surface is parallel to the axis, the angle of incidence at the wall is i′ = 90° − r. The condition i′ ≥ i_c gives a maximum value of r and hence a maximum angle of incidence at the end face: the acceptance angle. This is how fibres can carry light even when bent (NCERT Section 9.4.1, Fig. 9.14).

Given: n₁ = 1.68 (core); n₂ = 1.44 (covering).

(a) With the outer covering

1. Critical angle at the core–cladding surface: sin i_c = n₂/n₁ = 1.44/1.68 = 0.857 ⇒ i_c = 59.0°.
2. Condition for TIR: the angle of incidence at the wall i′ ≥ 59°. Since i′ = 90° − r (r is the angle of refraction at the end face), r ≤ 90° − 59° = 31° (more precisely sin r_max = cos i_c = √(1 − 0.735) = 0.515).
3. Snell’s law at the end face (air → core): sin i = n₁ sin r ⇒ sin i_max = 1.68 × 0.515 = 0.865 ⇒ i_max ≈ 60°. (General result: sin i_max = √(n₁² − n₂²) = √(2.822 − 2.074) = 0.865.)
4. Range: all rays with angles of incidence (with the axis) between 0° and 60° are totally reflected inside the pipe: 0 < i < 60°.

(b) Without the outer covering (core in air)

1. Now the core is surrounded by air, n₂ = 1: sin i_c = 1/1.68 = 0.595 ⇒ i_c = 36.5°; TIR requires i′ ≥ 36.5° ⇒ r ≤ 90° − 36.5° = 53.5°.
2. Maximum allowed: sin i_max = 1.68 × sin 53.5° = 1.68 × 0.804 = 1.35 > 1 — impossible. (Equivalently √(n₁² − 1) = √(2.822 − 1) = 1.35 > 1.)
3. Since sin i can never exceed 1, every ray that enters the end face (for any angle of incidence from 0° up to 90°) meets the wall at an angle greater than 36.5° and is totally internally reflected. So the answer is: all angles of incidence (0°–90°) give total reflection inside the pipe.

• Why use a cladding then? If bare fibres touch each other or get contaminated (dust, moisture) light would leak out/cross-talk between fibres; the cladding protects the core surface, avoids losses and defines a controlled acceptance angle. In practice a fibre with a very small difference n₁ − n₂ has a narrow acceptance angle (low dispersion) suited to long-distance communication.
• Bending: because the angles at the wall are large the light is guided even if the fibre is bent gently — as in endoscopes and optical-fibre communication.

Conclusion: (a) total internal reflection occurs for rays with angles of incidence from 0° to 60° with the axis; (b) without the outer covering all rays entering the end face (every angle of incidence up to 90°) are totally reflected.

Concept and context. To obtain a real image of an object on a screen with a convex lens, the distance D between the object and the screen must be at least 4f: D ≥ 4f. For a given D, solving the lens formula 1/v − 1/u = 1/f with u + v = D (distances in magnitude) gives a quadratic equation u² − Du + Df = 0 for the object distance; it has real solutions only if D² − 4Df ≥ 0 ⇔ f ≤ D/4. The limiting case D = 4f has a single position of the lens (object and image both at 2f, a magnification of −1). Hence the largest focal length is D/4.

Given: object (bulb) to screen (opposite wall) distance D = 3 m = 300 cm.

1. Let the lens be at distance u from the bulb and v from the wall: u + v = D, and 1/v + 1/u = 1/f in magnitudes (object real, image real).
2. Substituting v = D − u: 1/(D − u) + 1/u = 1/f ⇒ D/[u(D − u)] = 1/f ⇒ u² − D u + D f = 0.
3. Real roots exist only if D² − 4 D f ≥ 0 ⇒ f ≤ D/4.
4. Maximum focal length: f_max = D/4 = 300/4 = 75 cm = 0.75 m (with the lens exactly midway, u = v = 150 cm = 2f, m = −1).

f_max = D/4 = 75 cm

• Physical reason: a convex lens forms a real image only if the object is beyond its focus; the smallest separation between a real object and its real image is 4f (when both are at 2f). If the wall is nearer than 4f, even moving the lens cannot produce a sharp image.
• Check for f = 75 cm: 1/150 + 1/150 = 1/75 ✓.

Conclusion: the maximum possible focal length of the lens is 75 cm (D/4); for a longer focal length no real image could be obtained on the opposite wall.

Concept and context. When an object and a screen are separated by a distance D greater than 4f, a convex lens can form a sharp real image on the screen at two different positions of the lens. These positions are conjugate: if at the first position the object distance is u₁ and the image distance v₁, then at the second position the object distance is v₁ and the image distance u₁ (the principle of reversibility of the light path). The two positions give a magnified image (lens close to the object) and a diminished one (lens close to the screen). The method of measuring f from the separation d of the two lens positions is the displacement method, which does not need the position of the optical centre of the lens: f = (D² − d²)/(4D).

Given: D = 90 cm; separation of the two lens positions d = 20 cm.

1. Let the first lens position be at u₁ from the object and image distance v₁ = D − u₁. At the second position the lens is at distance v₁ from the object (the roles of u and v are exchanged), so the displacement is d = v₁ − u₁ = 20 cm.
2. Solving u₁ + v₁ = 90 and v₁ − u₁ = 20: v₁ = 55 cm, u₁ = 35 cm.
3. Lens formula (first position): 1/f = 1/v₁ + 1/u₁ (magnitudes) = 1/55 + 1/35 = (7 + 11)/385 = 18/385.
4. f = 385/18 = 21.4 cm.
5. Check with the formula: f = (D² − d²)/(4D) = (90² − 20²)/(4 × 90) = (8100 − 400)/360 = 7700/360 = 21.4 cm ✓.

f = (D² − d²)/(4D) ≈ 21.4 cm

• Relation between the sizes: if h₁ and h₂ are the image sizes at the two positions, the size of the object is h = √(h₁h₂) (since m₁ m₂ = 1).
• Check of D ≥ 4f: 4 × 21.4 = 85.6 cm < 90 cm ✓ — two positions are possible. If d = 0 the two positions coincide and f = D/4 (Exercise 9.18).

Conclusion: the focal length of the convex lens is ≈ 21.4 cm.

Concept and context. For two thin lenses in contact the effective focal length is given by 1/F = 1/f₁ + 1/f₂. When the lenses are separated by a distance d the formula becomes 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂), where F is measured from the “principal planes” of the system, not from the lenses themselves. To find where a parallel beam is brought to focus, or to find the image of a given object, we must trace the rays through each lens in turn: the image formed by the first lens serves as the object for the second (which may be a virtual object if the image of the first lens lies beyond the second), and the overall magnification is m = m₁ × m₂.

Given: f₁ = +30 cm (convex), f₂ = −20 cm (concave), separation d = 8.0 cm. (a) parallel beam; (b) object height 1.5 cm at u₁ = −40 cm from the convex lens.

(a) Effective focal length

1. Formula: 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂) = 1/30 − 1/20 − 8/[30 × (−20)] = −1/60 + 8/600 = (−10 + 8)/600 = −2/600 = −1/300.
2. F = −300 cm (a diverging system with an effective focal length of 3 m).
3. Beam incident on the convex lens first: L₁ would focus a parallel beam 30 cm beyond itself; this point is 30 − 8 = 22 cm beyond L₂ and acts as a virtual object for L₂ (u = +22 cm): 1/v = 1/f₂ + 1/u = −1/20 + 1/22 = −1/220 ⇒ v = −220 cm. The emergent beam appears to diverge from a point 220 cm from L₂ (on the incident side).
4. Beam incident on the concave lens first: L₂ makes the beam appear to diverge from a point 20 cm in front of L₂ (virtual image on the incident side), which is at a distance 20 + 8 = 28 cm from L₁: object distance u = −28 cm for L₁: 1/v = 1/30 − 1/28 = −1/420 ⇒ v = −420 cm. The emergent beam appears to diverge from a point 420 cm from L₁.

• Does the answer depend on the side of incidence? The effective focal length defined through the principal planes is the same, F = −300 cm, but the position of the focus relative to the lenses depends on the side: 220 cm from L₂ when light enters through the convex lens, and 420 cm from L₁ when it enters through the concave lens. So the distance of the (virtual) focus from the lens system is not the same in the two cases.
• Is the notion useful? Not very. For lenses in contact the effective focal length is a complete description (it applies from the common position of the lenses). For separated lenses the distances of the principal planes from the lenses are different for the two sides and so one still has to trace the image through each lens in turn (as in part (b)); the concept of one “effective focal length” is of limited practical value except for approximate estimates.

(b) Image of the 1.5 cm object

1. Lens L₁ (convex): u₁ = −40 cm, f₁ = +30 cm: 1/v₁ = 1/30 − 1/40 = (4 − 3)/120 = 1/120 ⇒ v₁ = +120 cm; m₁ = v₁/u₁ = 120/(−40) = −3 (real, inverted, 4.5 cm).
2. Lens L₂ (concave): the image I₁ is 120 − 8 = 112 cm beyond L₂, so it is a virtual object: u₂ = +112 cm, f₂ = −20 cm: 1/v₂ = 1/f₂ + 1/u₂ = −1/20 + 1/112 = −0.0411 ⇒ v₂ = −24.3 cm (on the object side: a virtual image); m₂ = v₂/u₂ = −24.3/112 = −0.217.
3. Total magnification: m = m₁ × m₂ = (−3)(−0.217) = +0.65.
4. Size of the final image: h′ = m h = 0.652 × 1.5 = 0.98 cm — erect (m positive), virtual and diminished, located 24.3 cm to the left of the concave lens (on the same side as the object).

F = −300 cm ; m = +0.65 ; image size = 0.98 cm

Conclusion: (a) the effective focal length is −300 cm; the position of the focus relative to the lenses depends on the side of incidence (220 cm from the concave lens for incidence on the convex lens, 420 cm from the convex lens for the other side), so the notion of a single effective focal length is of limited use for such a system; (b) the magnification is +0.65 and the image is 0.98 cm tall (erect, virtual).

Concept and context. In a prism of refracting angle A, a ray refracted at the first face at angle r₁ meets the second face at angle of incidence r₂, with r₁ + r₂ = A. The ray just suffers total internal reflection at the second face when r₂ equals the critical angle i_c of the material, sin i_c = 1/n. The corresponding angle of incidence i at the first face follows from Snell’s law sin i = n sin r₁. For rays with a larger angle of incidence the angle r₁ is larger and r₂ smaller than i_c, so the ray emerges; for smaller angles of incidence (r₁ smaller) r₂ > i_c and TIR occurs at the second face.

Given: A = 60°; n = 1.524.

1. Critical angle: sin i_c = 1/n = 1/1.524 = 0.6562 ⇒ i_c = 41.0°.
2. Condition for just TIR at the second face: r₂ = i_c = 41.0°.
3. Angle of refraction at the first face: r₁ = A − r₂ = 60° − 41.0° = 19.0°.
4. Snell’s law at the first face: sin i = n sin r₁ = 1.524 × sin 19.0° = 1.524 × 0.3256 = 0.4962 ⇒ i = 29.7° ≈ 30°.

i ≈ 30°

• Result: a ray incident at about 30° (or at smaller angles of incidence, i.e. r₁ smaller, r₂ larger) is totally internally reflected at the second face; for angles of incidence greater than 30° it emerges through the second face.
• Check: with i = 30°, sin r₁ = 0.5/1.524 = 0.328 ⇒ r₁ = 19.15°, r₂ = 40.85° ≈ i_c ✓.
• Application: total-reflecting prisms (glass of critical angle ~42°, A = 90°, 45°–45°–90°) use this behaviour in periscopes and binoculars.

Conclusion: the ray should be incident at an angle of about 30° on the first face so that it just suffers total internal reflection at the second face.

Concept and context. A simple microscope (magnifying glass) is a convex lens of short focal length used with the object placed within the focal length, giving a virtual, erect, magnified image (NCERT Fig. 9.23). Two different “magnifications” must be distinguished: the linear (transverse) magnification m = v/u — the ratio of image size to object size, the size of the image itself; and the angular magnification (magnifying power) M — the ratio of the angle subtended at the eye by the image to the angle subtended by the object when it is placed at the near point D = 25 cm (the least distance of distinct vision) without the lens. They are equal only when the image is formed at the near point (|v| = D). Areas scale as m². Note on the data: the printed exercise has f = 9 cm with the object at 9 cm; with the object exactly at the focus the image is at infinity and the linear magnification is not finite. The standard NCERT solution uses f = 10 cm (the value in the original book), which gives definite numerical answers — both are shown below.

Case A — data as printed (f = 9 cm, u = −9 cm)

• u = −f ⇒ 1/v = 1/f + 1/u = 1/9 − 1/9 = 0 ⇒ v = ∞: the image is formed at infinity, so the transverse magnification m = v/u is not finite and the “area of the image of each square” cannot be given. The angular magnification is however well defined: M = D/f = 25/9 ≈ 2.8 (image at infinity, relaxed eye).

Case B — standard NCERT solution (f = 10 cm, u = −9 cm)

1. Image distance: 1/v = 1/f + 1/u = 1/10 − 1/9 = (9 − 10)/90 = −1/90 ⇒ v = −90 cm (virtual image, 90 cm from the lens on the object side).
2. (a) Linear magnification: m = v/u = (−90)/(−9) = +10. Each linear dimension of a square is magnified 10 times; the area of each square in the virtual image is (10)² × 1 mm² = 100 mm² = 1 cm².
3. (b) Angular magnification (magnifying power): M = (angle subtended by the image)/(angle subtended by the object at the near point). The image subtends the same angle at the lens as the object does (h/9 cm for an object height h), while the unaided eye at 25 cm sees angle h/25: M = (h/9)/(h/25) = 25/9 = 2.8.
4. (c) Are they equal? No. The linear magnification (10) is a statement about the size of the image, while the magnifying power (2.8) compares the angles subtended at the eye. They are equal only when the image is formed at the near point D = 25 cm: |m| = |v|/|u| and M = D/|u| coincide only for |v| = D. Here the image is at 90 cm (farther than 25 cm), so the two differ; the 10-fold bigger image is much farther away and subtends only 2.8 times the angle.

Conclusion: (a) the linear magnification is 10 and each square of the image has an area of 100 mm² = 1 cm² (with f = 10 cm as in the standard solution; for the printed f = 9 cm the image is at infinity); (b) the magnifying power is 25/9 ≈ 2.8; (c) the two are not equal — they are equal only when the image is at the least distance of distinct vision (25 cm).

Concept and context. The magnifying power of a simple microscope is largest when the virtual image is formed at the near point (least distance of distinct vision) D = 25 cm: M_max = 1 + D/f. For the image at infinity it is only D/f. To view the squares distinctly the image must not be closer than 25 cm, so the maximum magnifying power is obtained with the image just at 25 cm; the lens must then be held at the object distance u found from the lens formula with v = −25 cm. (The data of Exercise 9.22 are used: f = 10 cm in the standard solution; the printed f = 9 cm is also shown.)

(a) Distance of the lens from the card

1. Image at the near point: v = −25 cm; f = +10 cm. 1/u = 1/v − 1/f = −1/25 − 1/10 = −(2 + 5)/50 = −7/50 ⇒ u = −50/7 = −7.14 cm.
2. The lens must be held about 7.1 cm from the card sheet. (For the printed f = 9 cm: 1/u = −1/25 − 1/9 = −34/225 ⇒ u = −6.62 cm.)

(b) Magnification in this case

1. m = v/u = (−25)/(−7.14) = 3.5 (for f = 9 cm: 25/6.62 = 3.78).
2. Equivalently m = 1 + D/f = 1 + 25/10 = 3.5 (1 + 25/9 = 3.78).

(c) Is the magnification equal to the magnifying power here?

• Yes. The image is at the near point (|v| = D = 25 cm), so the magnifying power M = (angle subtended by the image at 25 cm)/(angle subtended by the object at 25 cm) = (h′/25)/(h/25) = h′/h = m. When the image is at the near point, the linear magnification and the angular magnification coincide: M = m = 1 + D/f = 3.5.

u = −7.1 cm ; m = M = 1 + D/f = 3.5

Conclusion: (a) the lens should be held 7.1 cm from the card (6.6 cm for f = 9 cm); (b) the magnification is 3.5 (3.78 for f = 9 cm); (c) yes — when the image is at the near point (25 cm), the magnification equals the magnifying power, M = 1 + D/f.

Concept and context. The area of a (planar) object is magnified by the square of the linear magnification: area of image = m² × area of object. A virtual image of area 6.25 mm² for a square of 1 mm² requires m² = 6.25 ⇒ m = 2.5. The object distance for a given m follows from the lens formula (m = v/u). Whether the squares can be seen distinctly then depends on whether the virtual image lies at or beyond the near point (25 cm) of the eye. (The data of the exercises are used with f = 10 cm; the printed f = 9 cm gives a similar result.)

Given: object area 1 mm², image area 6.25 mm²; f = 10 cm (standard; for printed f = 9 cm see below).

1. Linear magnification: m² = 6.25/1 = 6.25 ⇒ m = 2.5 (the image is erect and virtual, so m = +2.5 = v/u).
2. Image distance in terms of object distance: v = 2.5 u (both negative: v = −2.5x if u = −x).
3. Lens formula: 1/v − 1/u = 1/f ⇒ −1/(2.5x) + 1/x = 1/10 ⇒ (1 − 0.4)/x = 1/10 ⇒ 0.6/x = 0.1 ⇒ x = 6 cm.
4. Hence the object (card) must be 6.0 cm from the lens, and the image distance is v = −2.5 × 6 = −15 cm.

Can the squares be seen distinctly?

• The virtual image is at 15 cm from the lens, i.e. (with the eye very close to the lens) at 15 cm from the eye — nearer than the near point of a normal eye (25 cm). The eye cannot focus on an object closer than 25 cm, so the image would look blurred: No, the squares cannot be seen distinctly. To see them clearly the image must be at a distance ≥ 25 cm (this requires the object to be moved to about 7.1 cm — Exercise 9.23).

• For the printed f = 9 cm: −1/(2.5x) + 1/x = 1/9 ⇒ 0.6/x = 1/9 ⇒ x = 5.4 cm and v = −13.5 cm — the same conclusion.
• Lesson (NCERT note): Exercises 9.22–9.24 clarify the difference between the magnification in absolute size (area ×6.25) and the angular magnification (the ability to see detail), which depends on the angle subtended at the eye and on the image being at or beyond the near point.

Conclusion: the object must be 6 cm from the lens (image at 15 cm); because the virtual image is closer than the near point (25 cm), the squares cannot be seen distinctly.

Concept and context. This exercise tests the concepts behind optical instruments. A simple microscope (magnifying glass) is a single convex lens used with the object inside its focal length; its angular magnification is M = 1 + D/f (D = 25 cm, the least distance of distinct vision) for the image at the near point. A compound microscope has an objective (short f_o) and an eyepiece (short f_e) with the total magnification M = (L/f_o)(D/f_e) approximately (L = distance between the second focus of the objective and the first focus of the eyepiece). The eye-ring (exit pupil) is the image of the objective formed by the eyepiece.

(a) In what sense does a magnifying glass give angular magnification?

• It is true that the angle subtended at the eye by the virtual image equals the angle subtended at the lens (and eye) by the object when the object is placed at the same position. The gain comes from the fact that the lens allows the object to be brought much closer to the eye than the near point (25 cm) — to a distance u < 25 cm (within the focal length) — while the eye still sees the virtual image comfortably at 25 cm or more. Without the lens the object could not be viewed distinctly at such a short distance. The angular size of the object at the near point (h/25) is smaller than the angular size of the image (h/u with u < 25); hence the magnifying glass provides angular magnification M = 25/u (= 1 + 25/f when the image is at 25 cm).

(b) Does the angular magnification change if the eye is moved back?

• Yes, it decreases (slightly). If the eye is moved away from the lens, the angle subtended at the eye by the image (which is at a fixed position) becomes smaller (the eye is farther from the image than the lens is, the image is seen further…), and also some of the rays leaving the lens may miss the pupil, reducing the field of view and the brightness. That is why one keeps the eye close to the lens.

(c) Why can we not use lenses of smaller and smaller focal length?

• Because of practical limitations: (i) very short focal length means very small radii of curvature, so the lens becomes extremely small and thick; it is difficult to grind and handle; (ii) aberrations (spherical and chromatic) increase rapidly for such strongly curved lenses, giving distorted and coloured images; (iii) the object must be very close to the lens, which is inconvenient and gives very little working distance and light; the object also becomes difficult to illuminate. In practice, simple microscopes achieve only magnifying powers of about 3 to 9; compound microscopes are used for higher magnification.

(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?

• The magnifying power M = m_o m_e ≈ (L/f_o)(D/f_e). For a given tube length L (which cannot be made arbitrarily large) and D = 25 cm, M is large only if both f_o and f_e are small. A short-focus objective makes the objective’s magnification (≈ L/f_o) large (the object is placed just outside f_o and the intermediate image is much larger); a short-focus eyepiece gives a large angular magnification (D/f_e). If f_o and f_e were large, the tube would have to be impracticably long for the same magnification.

(e) Why the eye should be placed a short distance from the eyepiece; how much?

• The best position for the eye is at the eye-ring (the exit pupil): the image of the objective (the aperture that limits the beam) formed by the eyepiece. All the light that has passed through the objective and the eyepiece passes through this small circle, so the eye placed there receives the maximum amount of light and has the maximum field of view. If the eye is placed on the eyepiece itself, the pupil is too close, and some of the light from the edges of the field falls outside the pupil; if it is too far, the light that has passed through the eye-ring spreads and the pupil collects only part of it.
• Distance: the eye-ring distance v is found from the lens formula for the eyepiece with the objective as the “object” at a distance equal to the separation L of the lenses: 1/v = 1/f_e − 1/L. For example, for the microscope of Exercise 9.12 (f_e = 2.5 cm, L = 9.47 cm): 1/v = 1/2.5 − 1/9.47 = 0.294 ⇒ v = 3.4 cm; for the microscope of Exercise 9.11 (f_e = 6.25 cm, L = 15 cm): v = 10.7 cm. In general, since L ≫ f_e, v is slightly larger than f_e: a short distance of a few centimetres (about the focal length of the eyepiece) behind the eyepiece.

Conclusion: (a) the magnification arises because the object can be viewed at less than 25 cm; (b) the angular magnification decreases slightly if the eye is moved back; (c) aberrations and practical difficulties limit the use of very short focal lengths; (d) M ∝ 1/(f_o f_e) for a limited tube length; (e) the eye should be placed at the eye-ring, a few centimetres from the eyepiece (about 3–4 cm for typical microscopes).

Concept and context. For a compound microscope the magnifying power is M = m_o × m_e, with the objective’s linear magnification m_o = v_o/u_o and the eyepiece’s angular magnification m_e = 1 + D/f_e (final image at the near point D = 25 cm). “Setting up” the microscope means choosing the object distance u_o and the distance between the lenses (L = v_o + |u_e|) such that the desired magnification is achieved. With f_o and f_e fixed, there is one required value of m_o. The final image is placed at the near point (maximum magnification, as in the standard solution); the alternative (image at infinity) is also given.

Given: M = 30; f_o = 1.25 cm; f_e = 5 cm; D = 25 cm.

Final image at the near point (25 cm)

1. Eyepiece magnification: m_e = 1 + D/f_e = 1 + 25/5 = 6.
2. Required objective magnification: |m_o| = M/m_e = 30/6 = 5 (m_o = −5, image inverted).
3. Object distance for the objective: m_o = v_o/u_o = −5 ⇒ v_o = 5|u_o|; with 1/v_o − 1/u_o = 1/f_o: 1/(5x) + 1/x = 1/1.25 (u_o = −x) ⇒ (1.2)/x = 0.8 ⇒ x = u_o = 1.5 cm; hence v_o = 7.5 cm.
4. Eyepiece position: the final image at v_e = −25 cm: 1/u_e = −1/25 − 1/5 = −6/25 ⇒ u_e = −4.17 cm. Distance between the lenses: L = v_o + |u_e| = 7.5 + 4.17 = 11.67 cm.

object 1.5 cm from the objective ; separation of the lenses = 11.67 cm

(Alternative: final image at infinity)

• Then m_e = D/f_e = 5, so |m_o| = 6: v_o = 6u_o; 1/(6x) + 1/x = 0.8 ⇒ x = 7/(6 × 0.8) = 1.46 cm and v_o = 8.75 cm; the lens separation = v_o + f_e = 8.75 + 5 = 13.75 cm.

Conclusion: place the object 1.5 cm from the objective and set the objective and eyepiece 11.67 cm apart (final image at 25 cm), giving a magnifying power of 30 (m_o = 5, m_e = 6).

Concept and context. The magnifying power of a refracting telescope is M = β/α, the ratio of the angle subtended at the eye by the final image to the angle subtended by the distant object. In normal adjustment (final image at infinity) the image formed by the objective lies at the focus of the eyepiece and M = f_o/f_e. If the eyepiece is moved so that the final image is formed at the near point D = 25 cm, the magnifying power increases to M = (f_o/f_e)(1 + f_e/D).

Given: f_o = 140 cm; f_e = 5.0 cm; D = 25 cm.

1. (a) Normal adjustment: M = f_o/f_e = 140/5.0 = 28.
2. (b) Final image at 25 cm: M = (f_o/f_e)(1 + f_e/D) = 28 × (1 + 5/25) = 28 × 1.2 = 33.6.

(a) M = 28 ; (b) M = 33.6

• Why larger in (b)? To form the final image at 25 cm the eyepiece is moved slightly closer to the objective image (u_e = 4.17 cm instead of 5 cm), so the eyepiece acts as a magnifier with M_e = 1 + D/f_e = 6 instead of D/f_e = 5: the factor (1 + f_e/D) = 1.2.
• Comfort: in normal adjustment the eye is relaxed (the emergent rays are parallel); viewing at 25 cm requires accommodation (strain) for a small gain.

Conclusion: (a) the magnifying power in normal adjustment is 28; (b) with the final image at the least distance of distinct vision it is 33.6.

Concept and context. In normal adjustment the telescope tube has a length equal to the sum of the focal lengths: L = f_o + f_e. The objective forms a real inverted image of the distant tower in its focal plane; the image height is h′ = f_o × (angle subtended by the tower) = f_o (h/d). The eyepiece then acts as a simple magnifier on this image: if the final image is at the near point D = 25 cm, the final image height is h″ = m_e h′ with m_e = 1 + D/f_e.

Given: f_o = 140 cm; f_e = 5.0 cm; tower height h = 100 m at a distance d = 3 km = 3000 m; D = 25 cm.

1. (a) Separation (normal adjustment): L = f_o + f_e = 140 + 5.0 = 145 cm.
2. (b) Image formed by the objective: angle subtended by the tower α = h/d = 100/3000 = 1/30 rad; the image is in the focal plane: h′ = f_o α = 140 × (1/30) = 4.67 cm (real, inverted). (Equivalent: m_o = v/u = f_o/d = 140/300000 and h′ = 100 m × 140/300000 cm… = 4.67 cm.)
3. (c) Final image at 25 cm: the eyepiece magnification m_e = 1 + D/f_e = 1 + 25/5 = 6; final image height h″ = m_e h′ = 6 × 4.67 = 28 cm (virtual, inverted relative to the tower).

• Check with the magnifying power: the angle subtended by the final image at the eye is β = h″/25 = 28/25 = 1.12 rad; M = β/α = 1.12 × 30 = 33.6 ✓ (matches Exercise 9.27(b)).
• The inverted image is the usual feature of an astronomical telescope; for terrestrial viewing an erecting system (extra lens or prisms) is added.

Conclusion: (a) the lenses are 145 cm apart; (b) the objective forms an image of the tower 4.67 cm tall; (c) the final image formed at 25 cm is 28 cm tall.

Concept and context. A Cassegrain reflecting telescope (NCERT Fig. 9.26) has a large concave primary mirror with a hole at its centre and a small convex secondary mirror facing it. Parallel light from a distant object is reflected by the primary towards its focus; before reaching the focus it meets the secondary, which reflects it back through the hole in the primary where the final image is formed (and viewed with an eyepiece). The image formed by the primary acts as the (virtual) object for the secondary. The design gives a long effective focal length in a short tube and, because the light is folded, the observer does not obstruct the incoming light (as in the prime-focus design of the Mt. Palomar telescope).

Given: distance between the mirrors d = 20 mm; R₁ = 220 mm (primary, concave); R₂ = 140 mm (secondary, convex).

1. Focal lengths: f₁ = R₁/2 = 110 mm; f₂ = R₂/2 = 70 mm.
2. Image formed by the primary of an object at infinity: at its focus, 110 mm from the primary. This image acts as a (virtual) object for the secondary mirror, which is 20 mm in front of the primary, so the virtual object distance is u = 110 − 20 = +90 mm (virtual, beyond the secondary mirror).
3. Mirror formula for the secondary: 1/v + 1/u = 1/f₂ ⇒ 1/v = 1/70 − 1/90 = (9 − 7)/630 = 2/630 = 1/315.
4. v = 315 mm. The final image is therefore formed 315 mm from the secondary mirror.

final image at v = 315 mm from the secondary mirror

• Remark: the Cassegrain arrangement gives an effective focal length much larger than that of the primary alone (here, the secondary magnifies the primary image by v/u = 315/90 = 3.5 in size), in a tube only 20 mm long between the mirrors — “a large focal length in a short telescope”.
• Largest telescopes: the 2.34 m Cassegrain reflector at Kavalur, Tamil Nadu (the largest in India) and the 10 m Keck telescopes (largest in the world), as mentioned in NCERT.

Conclusion: the final image is formed 315 mm from the secondary (small) mirror.

Concept and context. By the law of reflection the angle of reflection equals the angle of incidence. When a plane mirror is rotated through an angle θ about an axis in its plane (keeping the incident ray fixed), the normal rotates through θ, so the angle of incidence changes by θ and the reflected ray is rotated through 2θ. This “optical lever” doubles the effective deflection, and is used in sensitive galvanometers to magnify small rotations of the coil: the spot of light on a distant scale moves a distance d = L tan 2θ, where L is the distance of the scale from the mirror.

Given: deflection of the mirror θ = 3.5°; screen distance L = 1.5 m.

1. Rotation of the reflected ray: the incident ray is fixed and the mirror turns by θ = 3.5°, so the angle of incidence becomes 3.5° and the angle of reflection also 3.5°: the reflected ray makes an angle 2θ = 7° with the original (retraced) direction.
2. Displacement of the spot: d = L tan 2θ = 1.5 × tan 7° = 1.5 × 0.1228 = 0.184 m.
3. d ≈ 18.4 cm.

d = L tan 2θ = 1.5 tan 7° ≈ 0.18 m (18.4 cm)

• Optical lever advantage: a tiny rotation of the coil is magnified (doubled in angle and then multiplied by the long lever arm L): a 3.5° twist moves the light spot by 18 cm — a mechanical pointer of the same length would be heavy and affect the coil’s motion.
• Why 2θ? The normal rotates by θ, so both the incident-side and reflected-side angles with the normal change by θ; the reflected ray rotates 2θ relative to the fixed incident ray.

Conclusion: the reflected spot of light is displaced by about 18.4 cm on the screen, because the reflected ray turns through 2 × 3.5° = 7°.

Concept and context. When the needle’s image coincides with the needle (inverted, at the same position), the rays from the needle tip must strike the plane mirror normally and retrace their path: so the lens (and liquid layer) combination converts the diverging beam from the needle into a parallel beam, i.e. the needle is at the focal point of the lens + liquid combination. Therefore the measured needle distance equals the focal length of the combination: 45.0 cm with the liquid and 30.0 cm for the glass lens alone (with air). The liquid layer between the lens’s lower convex surface and the plane mirror forms a plano-concave liquid lens whose focal length can be deduced from 1/f = 1/f₁ + 1/f₂ (lenses in contact); the liquid’s refractive index then follows from the lens maker’s formula.

Given: n_glass = 1.50; combination focal length (with liquid) f′ = 45.0 cm; focal length of the lens alone f = 30.0 cm.

1. Radius of curvature of the equiconvex lens: 1/f = (n − 1)(1/R + 1/R) = 2(n − 1)/R ⇒ 1/30 = 2 × 0.5/R ⇒ R = 30 cm.
2. Focal length of the liquid lens: the combination of the glass lens (f = 30 cm) and the liquid lens (f_L) has 1/f′ = 1/f + 1/f_L ⇒ 1/45 = 1/30 + 1/f_L ⇒ 1/f_L = 1/45 − 1/30 = (2 − 3)/90 = −1/90 ⇒ f_L = −90 cm (a diverging, plano-concave lens).
3. Lens maker’s formula for the liquid lens: its upper surface fits the lower surface of the glass lens: radius of curvature R₁ = −30 cm (the centre of curvature lies on the incident side) and its lower surface is plane (R₂ = ∞): 1/f_L = (n_L − 1)(1/R₁ − 1/R₂) = (n_L − 1)(−1/30).
4. Solve: −1/90 = −(n_L − 1)/30 ⇒ n_L − 1 = 30/90 = 1/3 ⇒ n_L = 1.33.

refractive index of the liquid n = 1.33

• Why does the image coincide with the needle? The needle at the focus gives a parallel beam after the lens; the plane mirror reflects it back parallel; the lens (and liquid) focus it again at the same point: an inverted image at the position of the needle (no parallax). This is a standard laboratory method for measuring f and refractive indices of liquids.
• The value 1.33 is the refractive index of water.

Conclusion: the refractive index of the liquid is 1.33 (the liquid behaves like a plano-concave lens of focal length −90 cm).