Ray Optics and Optical Instruments Class 12 Physics NCERT Solutions

Step-by-step solution

1. 1For a concave mirror, f = −R/2 = −36/2 = −18 cm; u = −27 cm.
2. 2Mirror formula 1/v + 1/u = 1/f gives 1/v = 1/f − 1/u = −1/18 + 1/27 = −1/54, so v = −54 cm.
3. 3The screen must be placed 54 cm in front of the mirror.
4. 4Magnification m = −v/u = −(−54)/(−27) = −2, so the image is real, inverted and enlarged.
5. 5Size h′ = m·h = 2 × 2.5 = 5 cm.
6. 6As the candle moves closer, the image moves further away and grows; the screen must be moved progressively farther from the mirror. (At u = f no image is formed.)

Step-by-step solution

1. 1For a convex mirror, f = +15 cm; u = −12 cm.
2. 21/v = 1/f − 1/u = 1/15 + 1/12 = 9/60, so v = +6.7 cm behind the mirror (virtual).
3. 3Magnification m = −v/u = −6.7/(−12) = 0.56.
4. 4Image height h′ = m·h = 0.56 × 4.5 = 2.5 cm, erect and diminished.
5. 5As the needle moves farther away, its image moves toward the focus (15 cm behind the mirror) and keeps shrinking — the virtual image always lies between the pole and the focus.

Step-by-step solution

1. 1Refractive index μ = apparent depth/real depth ... μ = real depth/apparent depth = 12.5/9.4 = 1.33.
2. 2For the new liquid (μ = 1.63), apparent depth = real depth/μ = 12.5/1.63 = 7.67 cm.
3. 3Earlier the microscope was focused at 9.4 cm; it now focuses at 7.67 cm below the surface.
4. 4The microscope must be raised by 9.4 − 7.67 = 1.73 cm.

Step-by-step solution

1. 1From Fig. 9.27(a): n_g sin r_g = n_air sin 60°, giving n_g = sin 60°/sin 35.3° ≈ 1.5.
2. 2From Fig. 9.27(b): n_w = sin 60°/sin 40.6° ≈ 1.33.
3. 3At the water-glass interface, Snell's law gives n_w sin 45° = n_g sin r.
4. 4sin r = (n_w/n_g) sin 45° = (1.33/1.5) × 0.7071 = 0.627.
5. 5r = sin⁻¹(0.627) = 38.8°.

Step-by-step solution

1. 1Light escapes only within the critical cone: sin i_c = 1/μ = 1/1.33 = 0.752, so i_c = 48.75°.
2. 2The emergent circle has radius r = h tan i_c = 80 × tan 48.75° = 80 × 1.14 = 91.2 cm.
3. 3Area = πr² = π × (0.912 m)² = 2.6 m².

Step-by-step solution

1. 1μ = sin[(A + δ_m)/2]/sin(A/2) = sin[(60° + 40°)/2]/sin 30° = sin 50°/0.5.
2. 2μ = 0.7660/0.5 = 1.532.
3. 3In water, the prism acts with a reduced relative index μ′ = μ/μ_w = 1.532/1.33 = 1.152.
4. 4sin[(A + δ′)/2] = μ′ sin(A/2) = 1.152 × 0.5 = 0.576, giving (A + δ′)/2 = 35.2°.
5. 5δ′ = 2 × 35.2° − 60° = 10.4° ≈ 10°.

Step-by-step solution

1. 1For an equiconvex lens in air, 1/f = (μ − 1)[1/R − 1/(−R)] = 2(μ − 1)/R.
2. 21/20 = 2(1.55 − 1)/R = 2 × 0.55/R = 1.1/R.
3. 3R = 1.1 × 20 = 22 cm.

Step-by-step solution

1. 1The converging beam forms a virtual object for the lens: u = +12 cm (behind the lens, in the direction of travel).
2. 2(a) Convex lens, f = +20 cm: 1/v = 1/f + 1/u = 1/20 + 1/12 = 8/60, so v = 7.5 cm.
3. 3The beam converges 7.5 cm in front of the lens (between lens and P).
4. 4(b) Concave lens, f = −16 cm: 1/v = −1/16 + 1/12 = 1/48, so v = 48 cm.
5. 5The beam now converges 48 cm from the lens, beyond P.

Step-by-step solution

1. 1Concave lens: f = −21 cm, u = −14 cm.
2. 21/v = 1/f + 1/u = −1/21 − 1/14 = −5/42, so v = −8.4 cm.
3. 3Image is virtual, formed 8.4 cm from the lens on the same side as the object.
4. 4Magnification m = v/u = (−8.4)/(−14) = 0.6; height = 0.6 × 3.0 = 1.8 cm, erect.
5. 5Moving the object further away moves the virtual image towards the focus (−21 cm), making it smaller still.

Step-by-step solution

1. 1For lenses in contact, 1/F = 1/f₁ + 1/f₂ = 1/30 + 1/(−20).
2. 21/F = (2 − 3)/60 = −1/60, so F = −60 cm.
3. 3A negative focal length means the combination is a diverging lens.

Step-by-step solution

1. 1(a) Eyepiece forming the final image at the near point: 1/f_e = 1/v_e − 1/u_e with v_e = −25 cm.
2. 21/u_e = 1/v_e − 1/f_e = −1/25 − 1/6.25 = −0.2, so u_e = −5 cm.
3. 3The objective image must be at v_o = 15 − 5 = 10 cm from the objective.
4. 41/f_o = 1/v_o − 1/u_o gives 1/u_o = 1/10 − 1/2 = −0.4, so u_o = −2.5 cm.
5. 5m_o = v_o/|u_o| = 10/2.5 = 4 and m_e = 1 + D/f_e = 1 + 25/6.25 = 5.
6. 6Magnifying power M = m_o × m_e = 4 × 5 = 20.
7. 7(b) Final image at infinity: u_e = f_e = 6.25 cm, so v_o = 15 − 6.25 = 8.75 cm.
8. 81/u_o = 1/8.75 − 1/2 = −0.386, so u_o = −2.59 cm.
9. 9M = (v_o/|u_o|)(D/f_e) = (8.75/2.59) × (25/6.25) = 3.38 × 4 = 13.5.

Step-by-step solution

1. 1Objective: u_o = −0.9 cm, f_o = 0.8 cm. 1/v_o = 1/f_o − 1/|u_o| ... use 1/v_o = 1/f_o + 1/u_o = 1/0.8 − 1/0.9.
2. 21/v_o = 1.25 − 1.111 = 0.139 cm⁻¹, so v_o = 7.2 cm.
3. 3Eyepiece for near-point viewing: v_e = −25 cm, f_e = 2.5 cm.
4. 41/u_e = −1/25 − 1/2.5 = −0.44, so u_e = −2.27 cm.
5. 5Separation of lenses L = v_o + |u_e| = 7.2 + 2.27 = 9.47 cm.
6. 6m_o = v_o/|u_o| = 7.2/0.9 = 8; m_e = 1 + D/f_e = 1 + 25/2.5 = 11.
7. 7M = m_o × m_e = 8 × 11 = 88.

Step-by-step solution

1. 1In normal adjustment, magnifying power M = f₀/fₑ = 144/6.0 = 24.
2. 2Separation of lenses L = f₀ + fₑ = 144 + 6.0 = 150 cm.

Step-by-step solution

1. 1(a) M = f₀/fₑ = 15 m/1.0 cm = 1500 cm/1.0 cm = 1500.
2. 2(b) Angular size of the moon θ = D_moon/d = 3.48 × 10⁶ / 3.8 × 10⁸ = 9.16 × 10⁻³ rad.
3. 3Image diameter = f₀ θ = 15 × 9.16 × 10⁻³ = 0.137 m = 13.7 cm.

Step-by-step solution

1. 1(a) Concave mirror: f = −f₁, u = −u₁ with f₁ < u₁ < 2f₁. From 1/v = 1/f − 1/u = −1/f₁ + 1/u₁ = (u₁ − f₁)/(f₁u₁), v = f₁u₁/(u₁ − f₁) > 0, so the image is real and in front of the mirror.
2. 2Since u₁ < 2f₁, v = f₁u₁/(u₁ − f₁) > 2f₁ ⇔ u₁ < 2f₁, which holds — hence the image lies beyond 2f.
3. 3(b) Convex mirror: f = +f₁, u = −u₁. Then 1/v = 1/f₁ + 1/u₁, so v = f₁u₁/(f₁ + u₁) < 0 — the image is always behind the mirror, i.e. virtual, whatever the object position.
4. 4(c) From (b), |v| = f₁u₁/(f₁ + u₁) < f₁, so the image lies between pole and focus. Also |m| = |v/u| = f₁/(f₁ + u₁) < 1, so the image is diminished.
5. 5(d) Concave mirror, object between pole and focus: u = −u₁ with u₁ < f₁. Then 1/v = −1/f₁ + 1/u₁ = (f₁ − u₁)/(f₁u₁) > 0, so v > 0 — a virtual image on the other side. |m| = |v/u| = f₁/(f₁ − u₁) > 1, so the image is enlarged.

Step-by-step solution

1. 1Normal shift produced by a slab: Δ = t(1 − 1/μ) = 15(1 − 1/1.5) = 15 × 1/3 = 5 cm.
2. 2The pin appears raised by 5 cm.
3. 3The shift depends only on the thickness and refractive index of the slab, not on where the slab is held along the line of sight.

Step-by-step solution

1. 1Critical angle at the core-cladding interface: sin i_c = n₂/n₁ = 1.44/1.68 = 0.857, i_c = 59°.
2. 2Inside the core the ray must meet the cladding at r > i_c, i.e. angle with the axis α < 90° − i_c ≈ 31°.
3. 3From Snell's law at entry from air: sin i = n₁ sin α_max = 1.68 sin 31° = 0.87, giving i_max ≈ 60°.
4. 4So rays within ±60° of the axis are guided.
5. 5(b) Without cladding, n₂ = 1: sin i_max = √(n₁² − n₂²) = √(1.68² − 1) = 1.35 > 1, so totally internally reflected for all angles up to 90°.

Step-by-step solution

1. 1Object and screen are fixed 3 m apart: u + v = 3 m.
2. 2For a real image the object and image distances must satisfy v ≥ 4f (minimum distance between a real object and its real image is 4f).
3. 3With u = v = 1.5 m, the lens formula 1/f = 1/v + 1/|u| gives 1/f = 2/1.5.
4. 4f_max = 1.5/2 = 0.75 m.

Step-by-step solution

1. 1In the displacement method, D = 90 cm and d = 20 cm.
2. 2f = (D² − d²)/4D = (90² − 20²)/(4 × 90).
3. 3f = (8100 − 400)/360 = 7700/360 = 21.4 cm.

Step-by-step solution

1. 1(a) Convex lens f₁ = +30 cm, concave lens f₂ = −20 cm, separated by d = 8 cm.
2. 2Equivalent power 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂) = 1/30 − 1/20 − 8/(30 × −20).
3. 31/F = 0.0333 − 0.05 + 0.0133 = −0.0033, so F = −300 cm (diverging).
4. 4The answer depends on the side of incidence; a single 'effective focal length' is not a complete description (the principal planes matter), but F = −300 cm is the standard result given.
5. 5(b) For the first lens: u₁ = −40 cm, f₁ = +30 cm → 1/v₁ = 1/30 − 1/40 = 1/120, so v₁ = 120 cm.
6. 6This image is 120 − 8 = 112 cm beyond the second lens — a virtual object for it: u₂ = +112 cm.
7. 71/v₂ = 1/f₂ + 1/u₂ = −1/20 + 1/112 = −0.041, so v₂ = −24.35 cm (final image 24.35 cm to the left of lens 2).
8. 8m₁ = v₁/u₁ = 120/(−40) = −3; m₂ = v₂/u₂ = −24.35/112 = −0.217.
9. 9Total m = m₁m₂ = 0.652; image size = 0.652 × 1.5 = 0.98 cm, virtual and erect.

Step-by-step solution

1. 1Critical angle in the prism: sin i_c = 1/μ = 1/1.524 = 0.656, so i_c = 41°.
2. 2For just-total internal reflection the ray meets the second face at r₂ = i_c = 41°.
3. 3Geometry of the prism: r₁ + r₂ = A = 60°, so r₁ = 60° − 41° = 19°.
4. 4Snell's law at the first face: sin i = μ sin r₁ = 1.524 × sin 19° = 1.524 × 0.3256 = 0.496.
5. 5i = sin⁻¹(0.496) = 29.7° ≈ 29°45′.

Step-by-step solution

1. 1The printed 'focal length 9 cm' is a known typographical error in this edition; the consistent set of answers (including Exercises 9.23 and 9.24) uses the intended f = 10 cm with the card at u = −9 cm.
2. 2(a) 1/v = 1/f + 1/u = 1/10 − 1/9 = −1/90, so v = −90 cm (virtual image 90 cm behind the lens).
3. 3Magnification m = |v/u| = 90/9 = 10.
4. 4Each 1 mm × 1 mm square becomes 10 mm × 10 mm, area = 100 mm² = 1 cm².
5. 5(b) Angular magnification (magnifying power) = D/|u| = 25/9 = 2.8.
6. 6(c) No. The linear magnification m = |v/u| = 10 differs from the angular magnifying power α = β/α = D/|u| = 2.8; they coincide only when the image is formed at the near point D.

Step-by-step solution

1. 1(a) Maximum magnifying power requires the final image at the near point: v = −25 cm, f = +10 cm (intended value, as in Exercise 9.22).
2. 22. 1/u = 1/v − 1/f = −1/25 − 1/10 = −7/50, so u = −7.14 cm.
3. 3The lens must be held 7.14 cm from the card.
4. 4(b) Magnification = |v/u| = 25/7.14 = 3.5.
5. 5(c) Yes — with the image exactly at the near point, magnifying power D/|u| = 25/7.14 = 3.5 equals the linear magnification.

Step-by-step solution

1. 1Area 6.25 mm² means each side is √6.25 = 2.5 mm, so the linear magnification m = 2.5.
2. 2For the magnifier, m = |v/u| = 2.5 with f = +10 cm (intended value).
3. 3With u = −x, v = −2.5x: 1/f = 1/v − 1/u = −1/(2.5x) + 1/x = 0.6/x.
4. 41/10 = 0.6/x, so x = 6 cm. The card must be 6 cm from the lens, and |v| = 15 cm.
5. 5The image forms 15 cm from the eye, well inside the near point (25 cm) — the squares cannot be seen distinctly.

Step-by-step solution

1. 1(a) Without the lens the object can be brought no closer than the near point D ≈ 25 cm, so it subtends a small angle. The magnifier lets the object come close to the eye (where its angle is large) while the lens bends the rays so the eye sees a clear virtual image — the angular magnification is the ratio of the larger angle (with the lens) to the smaller angle available without it.
2. 2(b) Slightly, in principle — the angular size of the virtual image shrinks as the eye moves back, so magnification decreases; with the image effectively at infinity the change is negligible, which is why the eye is held close.
3. 3(c) Practical limits: aberrations (spherical and chromatic) grow with curvature, the lens becomes impractically small and thick, and the object would have to sit inside a very small focal region. In practice a simple microscope tops out around 9×.
4. 4(d) Magnifying power M = m_o m_e with m_e = 1 + D/f_e; short f₀ and fₑ keep the lengths small and make v₀/u₀ and D/fₑ large, giving a big total magnification in a compact tube.
5. 5(e) The eye should be placed at the exit pupil — the image of the aperture formed just beyond the eyepiece — where the whole emergent cone of rays can enter the eye (largest brightness and field). That distance is a few mm outside the eyepiece.

Step-by-step solution

1. 1Choose near-point (v_e = D = 25 cm) viewing for the best magnifying power: m_e = 1 + D/f_e = 1 + 25/5 = 6.
2. 2Required objective magnification m_o = M/m_e = 30/6 = 5 = v_o/|u_o|.
3. 3With u_o = −x, v_o = 5x and 1/f_o = 1/v_o − 1/u_o = 1/(5x) + 1/x = 1.2/x.
4. 41/1.25 = 1.2/x → x = 1.5 cm, so u_o = −1.5 cm and v_o = 7.5 cm.
5. 5For the eyepiece: 1/u_e = 1/v_e − 1/f_e = −1/25 − 1/5 = −0.24, so u_e = −4.17 cm.
6. 6Separate the lenses by L = v_o + |u_e| = 7.5 + 4.17 = 11.7 cm; place the object 1.5 cm from the objective.

Step-by-step solution

1. 1(a) Normal adjustment: M = f₀/fₑ = 140/5.0 = 28.
2. 2(b) Final image at the near point: M = f₀/fₑ (1 + fₑ/D) = 28 × (1 + 5/25).
3. 3M = 28 × 1.2 = 33.6.

Step-by-step solution

1. 1(a) Separation in normal adjustment = f₀ + fₑ = 140 + 5.0 = 145 cm.
2. 2(b) Angular size of the tower: θ = 100/3000 = 1/30 rad.
3. 3Image height at the objective focus h = f₀ θ = 140 × (1/30) = 4.7 cm.
4. 4(c) The eyepiece magnifies this by m_e = 1 + D/fₑ = 1 + 25/5 = 6.
5. 5Final image height = 6 × 4.7 = 28 cm.

Step-by-step solution

1. 1Focal length of the large (concave) mirror: f₁ = R₁/2 = 220/2 = 110 mm.
2. 2Rays from infinity focus 110 mm in front of the large mirror, i.e. 110 − 20 = 90 mm beyond the small mirror.
3. 3This focus acts as a virtual object for the small (convex) mirror: f₂ = R₂/2 = 70 mm and u = 90 mm.
4. 4Mirror formula for the small mirror: 1/v + 1/u = 1/f₂ → 1/v = 1/70 − 1/90 = 2/630.
5. 5v = 315 mm — the final image is formed 315 mm from the small mirror.

Step-by-step solution

1. 1When the mirror turns through φ, the reflected ray turns through 2φ = 7°.
2. 2On a screen at distance L = 1.5 m, displacement d = L tan 7°.
3. 3d = 1.5 × 0.1228 = 0.184 m = 18.4 cm.

Step-by-step solution

1. 1When the image coincides with the needle, the object is at the focal point of the equivalent optical system: combined focal length f₁ = 45.0 cm, glass-lens-only focal length f₂ = 30.0 cm.
2. 2Combination of lenses in contact: 1/f₁ = 1/f₂ + 1/f_liquid → 1/f_liquid = 1/45 − 1/30 = −1/90, so f_liquid = −90 cm.
3. 3For the equiconvex glass lens: 1/f₂ = (μ_g − 1)(1/R + 1/R) = 2(0.5)/R = 1/R, so R = 30 cm.
4. 4The liquid layer forms a plano-concave lens: 1/f_liquid = (μ_l − 1)(1/R₁ − 1/R₂) with R₁ = −R = −30 cm (curved face) and R₂ = ∞.
5. 5−1/90 = (μ_l − 1)(−1/30) → μ_l − 1 = 1/3, so μ_l = 4/3 = 1.33.