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

What is being asked. Find the image position for a candle placed in front of a concave mirror, and check whether the image is real (catchable on a screen) or virtual, magnified or diminished, erect or inverted.

Given.

• Object height, h₁ = +2.5 cm (above the axis, so positive)
• Object distance, u = -27 cm (object is in front of the mirror — negative in the Cartesian convention)
• Radius of curvature, R = -36 cm (centre of curvature is on the same side as the object for a concave mirror — negative)
• Focal length, f = R/2 = -18 cm

Concept used — mirror formula and magnification. 1v+1u = 1f, m = h₂h₁ = - vu. The mirror formula relates the image distance v, object distance u and focal length f. Magnification m tells us how tall the image is and (by its sign) whether it is inverted (m<0) or erect (m>0).

Step 1 — rearrange for v. 1v = 1f-1u.

Step 2 — substitute the numbers. 1v = 1-18-1-27 = -118+127.

Step 3 — common denominator (LCM of 18 and 27 is 54). 1v = -354+254 = -154.

Step 4 — invert. v = -54 cm. The minus sign means the image is on the same side as the object (in front of the mirror) — so it is REAL and can be caught on a screen 54 cm from the mirror.

Step 5 — magnification. m = -vu = -(-54)(-27) = -2. The negative sign means the image is inverted; the magnitude 2 means the image is twice as tall as the object.

Step 6 — image height. h₂ = m h₁ = (-2)(2.5) = -5 cm. So the image is 5 cm tall, inverted (below the axis).

Step 7 — what if the candle moves closer to the mirror? Move it from 27 cm toward the focus at 18 cm. As u → -f, the mirror formula gives v → ∞. So the screen must be moved farther and farther away to catch the image; once the candle is inside the focus (|u| < 18 cm), no real image forms — the image becomes virtual and the screen cannot catch it at all.

Final answer. v = -54 cm (real, in front of mirror), h₂ = -5 cm (inverted, magnified 2×). Screen at 54 cm from the mirror. As the candle approaches the focus, the screen must be moved away to infinity.

Given.

• Object height, h₁ = +4.5 cm
• Object distance, u = -12 cm
• Focal length, f = +15 cm (convex mirror — positive)

Concept used — mirror formula. 1v+1u=1f, m=-vu=h₂h₁. A convex mirror has a virtual focus behind it; light only appears to diverge from there. Such a mirror always produces a virtual, erect, diminished image for a real object — but we'll prove it from the formula.

Step 1 — solve for v. 1v=1f-1u=115-1-12=115+112.

Step 2 — common denominator (LCM 60). 1v=460+560=960=320.

Step 3 — invert. v=203 cm ≈ +6.7 cm. The positive sign means the image is behind the mirror — so it is VIRTUAL and erect; a screen cannot catch it.

Step 4 — magnification. m=-vu=-20/3-12=2036=59 ≈ 0.56.

Step 5 — image height. h₂ = m h₁ = 59× 4.5 = 2.5 cm. The positive sign confirms the image is erect; the magnitude 0.56 means it is diminished — about half the size.

Step 6 — as the needle moves away. Take u→ -∞. The mirror formula gives 1/v → 1/f so v→ +f = +15 cm. The image moves toward the focus behind the mirror, but never reaches it, while m → 0 — the image becomes a tiny dot at the focus.

Final answer. v=+6.7 cm (virtual, behind mirror), m=0.56, h₂ = 2.5 cm (erect, diminished). As the needle moves away, the image shrinks and recedes toward the focus, never crossing it.

Setup. A needle at the bottom of a tank looks closer to the top surface than it really is, because rays bend (refract) when they leave the water into air. We use the "apparent depth" relation.

Given.

• Real depth, d_real=12.5 cm
• Apparent depth in water, d_app=9.4 cm
• For part (b): replaced liquid has μ' = 1.63

Concept used — apparent depth. When looking nearly straight down through a transparent medium of refractive index μ into air, μ = real depthapparent depth. The microscope is focused at the apparent depth, where the rays seem to come from.

Step 1 — refractive index of water. _w = 12.59.4 ≈ 1.33. This matches the textbook value for water _w ≈ 1.33. Good check.

Step 2 — apparent depth in the new liquid. d_app' = d_realμ' = 12.51.63 ≈ 7.67 cm.

Step 3 — by how much must the microscope shift? Initially the microscope is focused at 9.4 cm (apparent depth in water). After replacing the water with the denser liquid, the needle appears at 7.67 cm — closer to the surface. The microscope must move down by Δ d = 9.4 - 7.67 = 1.73 cm ≈ 1.7 cm.

Final answer. _w ≈ 1.33, Δ d ≈ 1.73 cm (microscope moves toward the surface).

Reading the figure. From (a): in air → glass, incidence 60°, refraction 35°. From (b): in air → water, incidence 60°, refraction 47°. We need the angle of refraction in glass when light goes from water at 45°.

Concept used — Snell's law. ₁ sin₁ = ₂ sin₂. This holds at every interface. By chaining the refractive indices, we connect glass and water through air.

Step 1 — find _g (refractive index of glass relative to air) from Fig. (a). _ag = sin 60°sin 35° = 0.8660.574 ≈ 1.51.

Step 2 — find _w (refractive index of water relative to air) from Fig. (b). _aw = sin 60°sin 47° = 0.8660.731 ≈ 1.18. (NCERT uses these numbers; the standard "book" value for water is 1.33, but here we stick with what the figure gives us.)

Step 3 — relative refractive index of glass with respect to water. _wg = _agaw = 1.511.18 ≈ 1.28.

Step 4 — apply Snell's law at the water-glass interface. _w sin 45° = _g sin_g, ⇒ sin_g = _wgsin 45° = 1_wg sin 45°.

Step 5 — substitute. sin_g = 11.28× 0.707 = 0.552.

Step 6 — invert sine. _g = sin^-1(0.552) ≈ 38.5° ≈ 38°41'.

Final answer. _g ≈ 38.5°.

Setup. A point source at the bottom emits rays in all directions. Each ray hits the water surface; only those striking at angles less than the critical angle _c escape into the air. Rays steeper than _c are totally internally reflected back. The escaping rays form a bright "cone of light" — and the cone illuminates a CIRCLE on the surface.

Concept used — critical angle. sin_c = 1μ = n_rarern_denser. At _c the refracted ray skims the surface refraction angle =90°; beyond it, total internal reflection.

Given. Depth h=80 cm=0.8 m, μ=1.33.

Step 1 — find the critical angle. sin_c=11.33=0.752, _c=sin^-1(0.752)≈ 48.75°.

Step 2 — geometry of the lit circle. The bulb sits at depth h. Draw the limiting ray: it leaves the bulb, hits the water surface at angle _c from the vertical (normal). The horizontal distance from directly-above-the-bulb to the point it strikes the surface is the radius r of the lit circle: tan_c = rh, r = h tan_c.

Step 3 — compute tan_c. Use sin_c = 0.752, so cos_c = √1-0.752²=√1-0.566=√0.434=0.659. Therefore tan_c = 0.7520.659=1.141.

Step 4 — radius of the lit circle. r = 0.8 × 1.141 = 0.913 m.

Step 5 — area of the lit circle. A = π r² = π × (0.913)² = 3.1416× 0.833 = 2.62 m².

Final answer. A ≈ 2.62 m². The bulb illuminates a circular patch of about 2.6 m² on the water surface; everything outside the patch is dark above.

Given.

• Refracting angle of prism, A = 60°
• Angle of minimum deviation in air, D_m = 40°
• Refractive index of water, _w = 1.33

Concept used — prism formula at minimum deviation. At minimum deviation, light passes symmetrically through the prism, and μ = sin((A+D_m)/2)sin(A/2). This is the standard prism formula. Here μ is the refractive index of the prism relative to the surrounding medium.

Step 1 — find _g (prism in air). _g = sin((60+40)/2)sin(60/2) = sin 50°sin 30°.

Step 2 — substitute numerical values. sin 50° = 0.766, sin 30° = 0.5. _g = 0.7660.5 = 1.532. So the prism glass has refractive index ≈ 1.53 relative to air.

Step 3 — prism in water. Now light travels through water and into glass. The relevant refractive index is glass-relative-to-water: _wg = _gw = 1.5321.33 ≈ 1.152.

Step 4 — re-arrange prism formula for new D_m'. _wg = sin((A+D_m')/2)sin(A/2) ⇒ sin((A+D_m')/2) = _wg sin(A/2).

Step 5 — substitute. sin((60+D_m')/2) = 1.152 × sin 30° = 1.152 × 0.5 = 0.576.

Step 6 — invert sine. (60+D_m')/2 = sin^-1(0.576) = 35.17°.

Step 7 — solve for D_m'. 60 + D_m' = 70.34°, D_m' = 10.34° ≈ 10°.

Final answer. _g ≈ 1.53, D_m' (in water)≈ 10°. The prism deviates light much less when immersed in water, because the glass-water optical contrast is smaller than glass-air contrast.

Concept used — lensmaker's equation. For a thin lens in air with refractive index μ, and surfaces of radii R₁, R₂, 1f = (μ-1)(1R₁-1R₂). Sign convention: R is positive if the centre of curvature is on the outgoing (right) side, negative if on the incoming (left) side.

Given. μ = 1.55, f = +20 cm (converging double-convex lens).

Step 1 — assign signs to R₁ and R₂. For a double-convex (biconvex) lens with light incident from the left:
– Surface 1 bulges to the left (its centre of curvature is on the right) → R₁ = +R.
– Surface 2 bulges to the right (its centre of curvature is on the left) → R₂ = -R.
The problem says both faces have the same radius of curvature; we'll call its magnitude R.

Step 2 — plug into lensmaker's formula. 1f = (μ-1)(1+R - 1-R) = (μ-1)(1R+1R) = 2(μ-1)R.

Step 3 — solve for R. R = 2(μ-1) f.

Step 4 — substitute numbers. R = 2 (1.55-1) (20) = 2 × 0.55 × 20 = 22 cm.

Final answer. R = 22 cm. Each surface must be ground to a radius of curvature of 22 cm.

The trick — virtual object. Light is already converging toward P when it meets the lens. So for the lens, P acts as a virtual object on the far side. In the Cartesian convention, virtual object means u > 0 — measured along the direction of light propagation, on the outgoing side.

(a) Convex lens, f = +20 cm.

Step 1 — set up. Virtual object at u = +12 cm. Use thin-lens formula 1/v - 1/u = 1/f.

Step 2 — substitute. 1v = 1f + 1u = 120 + 112.

Step 3 — common denominator (LCM 60). 1v = 360 + 560 = 860 = 215.

Step 4 — invert. v = 152 = 7.5 cm. Positive, on the same side as P. So with the convex lens, the beam converges 7.5 cm past the lens (closer than P).

(b) Concave lens, f = -16 cm.

Step 1 — substitute. 1v = 1-16 + 112 = -116 + 112.

Step 2 — common denominator (LCM 48). 1v = -348 + 448 = 148.

Step 3 — invert. v = +48 cm. With the concave lens, the beam converges 48 cm past the lens (farther than P) — the lens has weakened the convergence.

Final answer. (a) v = +7.5 cm, (b) v = +48 cm.

Given.

• Object height, h₁ = +3.0 cm
• Object distance, u = -14 cm
• Focal length, f = -21 cm (concave lens → diverging → negative)

Concept used — thin-lens formula. 1v - 1u = 1f, m = vu = h₂h₁. For a concave lens, f < 0, and a real object (u < 0) always produces a virtual, erect, diminished image — let's prove it.

Step 1 — solve for v. 1v = 1f + 1u = 1-21 + 1-14 = -121 - 114.

Step 2 — common denominator (LCM 42). 1v = -242 - 342 = -542.

Step 3 — invert. v = -425 = -8.4 cm. Negative — image is on the same side as the object, i.e., virtual, erect.

Step 4 — magnification. m = vu = -8.4-14 = 0.6. Positive (erect), < 1 (diminished).

Step 5 — image size. h₂ = m h₁ = 0.6 × 3.0 = 1.8 cm.

Step 6 — as the object moves farther. Let u→ -∞. Then 1/v → 1/f = -1/21, so v → -21 cm; the image moves toward the focus on the object side. Meanwhile m→ 0 — image shrinks to zero size. So as object recedes, image moves toward focus and gets smaller, but stays virtual.

Final answer. v = -8.4 cm, h₂ = +1.8 cm (virtual, erect, diminished).

Concept used — combination of thin lenses in contact. When two thin lenses focal lengths f₁, f₂ are placed in contact, the combined focal length f is given by 1f = 1f₁ + 1f₂. Equivalently, in terms of powers P = 1/f in metres, P_net = P₁ + P₂.

Given. Convex lens: f₁ = +30 cm; concave lens: f₂ = -20 cm.

Step 1 — write the combination. 1f = 130 + 1-20 = 130 - 120.

Step 2 — common denominator (LCM 60). 1f = 260 - 360 = -160.

Step 3 — invert. f = -60 cm.

Step 4 — interpret. Negative focal length ⇒ the combination behaves as a diverging lens. So the concave lens "wins" — its power is greater than that of the convex lens, and the net system diverges.

Final answer. f = -60 cm, diverging system.

Concept used — compound microscope. A compound microscope has two converging lenses: objective (short focal length) forms a real, inverted, magnified image of the tiny object, which the eyepiece (acting as a simple magnifier) then magnifies further. The final image can be at infinity (relaxed eye) or at the near point D = 25 cm (maximum magnification).

Given. f_o = +2.0 cm, f_e = +6.25 cm, L = 15 cm (lens separation), D = 25 cm.

(a) Final image at near point v_e = -25 cm, erect virtual image relative to eyepiece.

Step 1 — locate the intermediate image (object for eyepiece). Apply lens formula to eyepiece: 1v_e - 1u_e = 1f_e ⇒ 1u_e = 1v_e - 1f_e = 1-25 - 16.25. LCM 25: 1u_e = -125 - 425 = -525 = -15. u_e = -5 cm.

Step 2 — locate the object for the objective. Distance from objective to intermediate image: v_o = L - |u_e| = 15 - 5 = 10 cm. So v_o = +10 cm. Apply lens formula to objective: 1u_o = 1v_o - 1f_o = 110 - 12. LCM 10: 1u_o = 110 - 510 = -410. So u_o = -2.5 cm.

Step 3 — magnification. m = m_o · m_e = v_ou_o(1 + Df_e) = 10-2.5(1+256.25) = -4 × (1+4) = -20. Magnitude |m|=20; the negative sign tells us the final image is inverted relative to the object.

(b) Final image at infinity (relaxed eye).

Step 1 — eyepiece. For final image at ∞, the intermediate image must be at the focus of the eyepiece, i.e., u_e = -f_e = -6.25 cm.

Step 2 — objective. v_o = L - 6.25 = 15 - 6.25 = 8.75 cm. 1u_o = 1v_o - 1f_o = 18.75 - 12 = 0.1143 - 0.5 = -0.3857. u_o = -10.3857 ≈ -2.59 cm.

Step 3 — magnification. m = v_ou_o · Df_e = 8.75-2.59 × 256.25 = -3.38 × 4 = -13.5.

Final answer.

• (a) Object at 2.5 cm in front of objective; magnifying power |m| = 20.
• (b) Object at ≈ 2.59 cm in front of objective; magnifying power |m| ≈ 13.5.

Given.

• f_o = 8.0 mm = 0.8 cm, u_o = -9.0 mm = -0.9 cm
• f_e = 2.5 cm, D = 25 cm
• Assume final image at near point (most magnification, standard).

Step 1 — image distance for objective. 1v_o = 1f_o + 1u_o = 10.8 + 1-0.9 = 1.25 - 1.111 = 0.139. v_o = 10.139 = 7.2 cm. So the objective produces a real, inverted, magnified image 7.2 cm behind itself.

Step 2 — find object distance for eyepiece. Final image at v_e = -25 cm. 1u_e = 1v_e - 1f_e = 1-25 - 12.5 = -0.04 - 0.4 = -0.44. u_e = -10.44 = -2.273 cm.

Step 3 — separation between lenses. The intermediate image is at v_o = 7.2 cm behind objective and serves as the eyepiece's object at |u_e| = 2.273 cm in front of the eyepiece. Therefore lens separation L = v_o + |u_e| = 7.2 + 2.273 = 9.47 cm.

Step 4 — magnification of objective. m_o = v_ou_o = 7.2-0.9 = -8.

Step 5 — magnification of eyepiece. m_e = 1 + Df_e = 1 + 252.5 = 1+10 = 11.

Step 6 — total magnification. m = m_o m_e = -8 × 11 = -88.

Final answer. L ≈ 9.47 cm, |m| = 88.

Concept used — astronomical telescope in normal adjustment. Object at infinity, final image at infinity. The objective forms a real image at its second focal point; the eyepiece then refocuses it for parallel viewing. Magnifying power: |m| = f_of_e, L = f_o + f_e. The minus sign convention (image inverted) is implied; for magnifying-power calculations we use magnitudes.

Given. f_o = 144 cm, f_e = 6.0 cm.

Step 1 — magnifying power. |m| = 1446.0 = 24.

Step 2 — tube length. L = 144 + 6 = 150 cm.

Final answer. |m| = 24, L = 150 cm.

(a) Angular magnification. |m| = f_of_e = 15 m0.01 m = 1500 cm1 cm = 1500.

(b) Diameter of moon's image at the objective's focal plane.

Step 1 — angular size of the moon (in radians). Since the moon is very far, its angular size is θ = diameterdistance = 3.48× 10⁶3.8× 10⁸ = 9.158× 10^-3 rad.

Step 2 — linear size of image at focal plane. The objective lens forms an image at distance f_o behind itself; the image's linear size is d_image = f_o · θ = 15 × 9.158× 10^-3 = 0.1374 m = 13.74 cm.

Final answer. |m| = 1500; d_image ≈ 13.7 cm.

Mirror formula and magnification. 1v+1u=1f, m=-vu.

(a) Concave mirror, 2f < u < f (in magnitude), i.e., -2f < u < -f, with f < 0. Write u in the range (-2|f|, -|f|). From the formula 1/v = 1/f - 1/u. With f < 0 and u between -2|f| and -|f|:
at u = -|f|: 1/v = 1/f - 1/f = 0 ⇒ v → ∞;
at u = -2|f|: 1/v = 1/f - 1/(2f) = 1/(2f) ⇒ v = 2f = -2|f|. So as u varies from -|f| (object at focus) to -2|f| (object at centre of curvature), v ranges from -∞ to -2|f|. All values of v are negative ⇒ image is real (on same side as object); |v| > 2|f| ⇒ image is beyond 2f. Proved.

(b) Convex mirror — virtual image always. For a convex mirror f > 0. For a real object, u < 0. Then 1v = 1f - 1u = 1f_>0 - 1u_<0 = 1f + 1|u|. Both terms positive ⇒ 1/v > 0 ⇒ v > 0 ⇒ image behind the mirror ⇒ virtual. Always. Proved.

(c) Convex mirror — image always diminished and between F and P. Continuing from (b): 1/v = 1/f + 1/|u|. Since 1/|u| > 0, 1/v > 1/f, so 0 < v < f. Image is between pole and focus. Now magnification m = -v/u = v/|u|. And from 1/v = 1/f + 1/|u|, we get v < |u| (since 1/v > 1/|u|). Therefore m = v/|u| < 1 ⇒ image diminished. Proved.

(d) Concave mirror, object between pole and focus, i.e., 0 < |u| < |f|, so -|f| < u < 0. Then 1/u is a large negative number (more negative than 1/f, since |u| < |f|): 1v = 1f - 1u = -1|f| - 1u. With -1/u > 1/|f| (because |u|<|f|), we get 1/v > 0 ⇒ v > 0 ⇒ image is behind mirror — VIRTUAL. Now magnification: m = -v/u = -v· 1/u. Since v > 0 and u < 0, m > 0 (erect). Also |m| = v/|u|. From 1/v = 1/f - 1/u with |u| < |f|, |v| > |u|, so |m| > 1. Image is enlarged and erect. Proved.

Final answer. All four properties follow directly from the mirror formula with the Cartesian sign convention.

Setup. A pin lies on the table; you look down at it through a glass slab in between. Refraction at the upper and lower faces of the slab makes the pin appear closer (raised) than it really is.

Concept used — apparent shift due to a slab. For viewing nearly along the normal, the apparent shift caused by a slab of thickness t and refractive index μ is Δ = t(1 - 1μ). This is independent of where the slab is placed in the optical path — only t and μ matter.

Given. t = 15 cm, μ = 1.5.

Step 1 — substitute. Δ = 15(1 - 11.5) = 15(1 - 23) = 15 × 13 = 5 cm.

Step 2 — interpret. The pin appears raised by 5 cm — that is, it looks as if it sits 50 - 5 = 45 cm below the eye, even though it's really 50 cm below.

Step 3 — does the slab's location matter? The formula Δ = t1-1/μ contains no reference to where the slab is in between the eye and the pin. So the answer is the same whether the slab is at the table, halfway up, or near the eye. The shift depends only on the slab's thickness and refractive index, not on its position.

Final answer. Δ = 5 cm; independent of slab location.

Setup. A glass fibre (core) is surrounded by a cladding of lower refractive index. Light enters the end face from air, refracts into the core, then hits the core-cladding interface at some angle. For the light to be totally internally reflected (TIR) and propagate along the fibre, that angle must exceed the critical angle at the core-cladding interface.

Concept used — total internal reflection. sin_c = n_claddingn_core. At the core-cladding interface, light must hit at angle ≥ _c for TIR.

(a) With cladding n_core = 1.68, n_cladding = 1.44.

Step 1 — find critical angle. sin_c = 1.441.68 = 0.857, _c = sin^-1(0.857) = 59°.

Step 2 — relate _c (at side wall) to i (at end face, from axis). The angle of the ray with the axis inside the fibre is the complement of the angle of incidence on the side wall. So when the ray hits the side wall at angle _c from the normal, it makes angle 90° - _c with the axis. 90° - _c = 90° - 59° = 31°. Inside the fibre, the ray makes angle 31° with the axis.

Step 3 — find external angle i (from axis) using Snell's law at the end face. The end face is perpendicular to the axis. So a ray at angle i with the axis (outside) refracts to angle r with the axis (inside), where r = 31°. sin i = n_core sin r = 1.68 × sin 31° = 1.68 × 0.515 = 0.865. i = sin^-1(0.865) = 60°.

Step 4 — range of allowed angles. Any external angle i ≤ 60° (from the axis) will undergo TIR inside the fibre. Larger i means smaller refracted angle (with the axis), means larger angle on the side wall — still safe. So the allowed range is 0° ≤ i ≤ 60°.

(b) Without cladding core surrounded by air, n_cladding = 1.

Step 1 — new critical angle. sin_c = 11.68 = 0.595, _c = 36.5°.

Step 2 — angle with axis inside. 90° - 36.5° = 53.5°. The ray must make at most 53.5° with the axis inside the fibre.

Step 3 — at the end face. sin i = 1.68 sin 53.5° = 1.68 × 0.804 = 1.35. But sin i cannot exceed 1! This means any external angle of incidence (up to 90°) results in TIR. The fibre captures light from a full hemisphere.

Final answer.

• (a) With cladding: 0° ≤ i ≤ 60° from the axis.
• (b) Without cladding: any angle works (full hemisphere capture).

Setup. Object on one wall, image on the opposite wall — separation d = 3 m between object and image. We need to find the maximum focal length for which a real image of the bulb forms on the opposite wall.

Concept used. For a convex lens, a real image of a real object can be formed only if the object-image distance d satisfies d ≥ 4f. The minimum is achieved when u = v = 2f — i.e., the object is at twice the focal length, the image is also at twice the focal length, and m = -1 (image and object same size).

Step 1 — write the condition. d = |u| + v ≥ 4f.

Step 2 — derive this. Use lens formula 1/v - 1/u = 1/f. Let |u| = a, v = b. Then a + b = d and 1/b + 1/a = 1/f. From the AM-GM inequality (or treating a, b as roots of a quadratic), one gets the minimum of (a+b) for fixed f is 4f, achieved when a = b = 2f.

Step 3 — apply the condition. 4f ≤ 3 m ⇒ f ≤ 0.75 m = 75 cm.

Step 4 — interpret. The maximum focal length is f = 0.75 m. With this lens, the bulb at 1.5 m is imaged onto the opposite wall at 1.5 m — same size, inverted, real. Any longer f would require a greater object-image separation than 3 m.

Final answer. f_max = 0.75 m = 75 cm.

Setup — Bessel's method. An object and screen are at fixed separation d = 90 cm. A convex lens placed between them can be at two distinct positions separated by x = 20 cm where it forms a sharp image on the screen. Bessel's formula gives the focal length directly: f = d² - x²4d.

Step 1 — substitute. f = (90)² - (20)²4 × 90 = 8100 - 400360 = 7700360 = 21.39 cm ≈ 21.4 cm.

Step 2 — verify via direct positions. Let the two lens-to-object distances be u₁, u₂; image distances v₁, v₂. By symmetry of the lens formula in u and v (swapping object and image positions), u₁ = v₂ and u₂ = v₁. So:
u₁ + v₁ = d ⇒ u₁ + u₂ = d = 90 cm calling v₁ = u₂.
u₂ - u₁ = x = 20 cm (the lens shift).
Solving: u₁ = 35 cm, u₂ = 55 cm.
Lens formula at position 1: 1/v₁ - 1/(-35) = 1/f, with v₁ = 90 - 35 = 55 cm.
1/55 + 1/35 = 1/f. LCM 385: 7/385 + 11/385 = 18/385 = 1/f, so f = 385/18 = 21.4 cm. Matches.

Final answer. f ≈ 21.4 cm.

Recall from Q 9.10. Convex lens f₁ = +30 cm, concave lens f₂ = -20 cm, now separated by d = 8 cm.

(a) Effective focal length with parallel light incident on the convex lens.

Step 1 — image from convex lens. Parallel beam means u₁ = -∞, so the convex lens forms an image at its focus: v₁ = +f₁ = +30 cm.

Step 2 — this image serves as virtual object for concave lens. Lens 2 is 8 cm from lens 1. The image of lens 1 forms 30 cm beyond lens 1, so it's 30 - 8 = 22 cm beyond lens 2 — i.e., a virtual object at u₂ = +22 cm.

Step 3 — lens formula for lens 2. 1v₂ = 1f₂ + 1u₂ = 1-20 + 122 = -0.05 + 0.0455 = -0.00455. v₂ = -220 cm. Negative — image is 220 cm to the left of lens 2.

Step 4 — interpret as effective focal length. The combination would behave like a single lens of focal length f_eff placed at some location, focusing parallel light at v = f_eff. Here the "image of parallel light" forms 220 cm to the left of lens 2 — but a normal converging system would form an image to the right. The negative v₂ signals that this combination, viewed from the convex-lens side, acts as a diverging system with virtual focus at ≈ 220 cm.

Step 5 — reverse direction (parallel light on concave side). Same procedure: parallel light hits the concave lens first, u₁ = -∞, v₁ = f₂ = -20 cm virtual image 20 cm before the concave lens. Then this serves as object for the convex lens 8 cm away, with u₂ = -(8 + 20) = -28 cm. 1v₂ = 130 + 1-28 = 0.0333 - 0.0357 = -0.00238. v₂ = -420 cm. Different answer! Effective focal length depends on which side parallel light enters.

Step 6 — usefulness. Because f_eff depends on direction, it's not a unique single number — the system is asymmetric for finite lens separation. Concept loses much of its utility; one must work with each lens separately.

(b) Object 1.5 cm, placed 40 cm from convex lens.

Step 1 — convex lens. u₁ = -40, f₁ = +30. 1v₁ = 130 - 140 = 4-3120 = 1120. v₁ = +120 cm. Real image 120 cm beyond convex lens.

Step 2 — magnification at lens 1. m₁ = v₁u₁ = 120-40 = -3. Intermediate image is 3× the object, inverted.

Step 3 — object for lens 2. Lens 2 is 8 cm beyond lens 1. The intermediate image is at 120 cm beyond lens 1, i.e., 120 - 8 = 112 cm beyond lens 2. Virtual object: u₂ = +112 cm.

Step 4 — apply lens 2. 1v₂ = 1-20 + 1112 = -0.05 + 0.00893 = -0.0411. v₂ = -24.3 cm. Hmm — virtual image, but image is on the input side of lens 2. Let's verify by re-checking: more careful computation. 1v₂ = 1f₂ + 1u₂ = -120 + 1112. LCM 560: -28/560 + 5/560 = -23/560. So v₂ = -560/23 = -24.3 cm.

Step 5 — magnification at lens 2. m₂ = v₂u₂ = -24.3112 = -0.217.

Step 6 — total magnification. m = m₁ · m₂ = (-3)×(-0.217) = +0.65.

Step 7 — image size. h₂ = m · h₁ = 0.65 × 1.5 = 0.975 cm ≈ 0.98 cm.

Final answer.

• (a) Effective focal length depends on direction of light: 220 cm or 420 cm (virtual foci on the input side in each case). Concept of single "effective f" is not very useful for this system.
• (b) Magnification ≈ 0.65; image size ≈ 0.98 cm, erect (positive m).

Setup. A ray enters the prism through face 1, traverses through the glass, and hits face 2. We want the geometry such that the angle of incidence at face 2 equals the critical angle (so it just undergoes TIR).

Concept used. Inside the prism, the two refracted angles (at face 1 and face 2) sum to the prism angle: r₁ + r₂ = A. At face 2, "just TIR" means r₂ = _c, where sin_c = 1/μ.

Given. A = 60°, μ = 1.524.

Step 1 — critical angle. sin_c = 11.524 = 0.6562, _c = sin^-1(0.6562) = 41.0°.

Step 2 — find r₁. r₁ = A - r₂ = 60° - 41.0° = 19.0°.

Step 3 — find external angle of incidence using Snell's law at face 1. sin i = μ sin r₁ = 1.524 × sin 19° = 1.524 × 0.3256 = 0.496.

Step 4 — invert. i = sin^-1(0.496) = 29.75° ≈ 29.75°.

Final answer. i ≈ 29.75°. Any incidence angle smaller than this allows the ray to pass through face 2 (no TIR); any larger angle causes TIR at face 2.

Setup. Card at u = -9 cm, magnifier with f = +9 cm. Note that with the object literally at the focal point the image forms at infinity, which would leave the linear magnification undefined. The NCERT solution circumvents this technicality by treating the image as forming at the near point v = -25 cm (where the eye can actually focus), and using the lens formula to find the corresponding v/u ratio.

(a) Linear magnification m and image-square area.

Step 1 — apply the lens formula with v = -25 cm, u = -9 cm (as quoted in the problem). m = vu = -25-9 × (NCERT convention: take the image at the near point so the linear magnification is finite). NCERT's accepted answer is obtained by setting the image at the near point (so the eye can actually see it) and computing m = v/u consistently. With v = -25 cm but the object kept at u = -9 cm, the lens formula isn't strictly satisfied for f = 9 cm; NCERT therefore treats f as effectively 10 cm (the lens "rounded up") so that the formula 1/v - 1/u = 1/f does hold: 1/(-25) - 1/(-9) = -1/25 + 1/9 = 16/225, close to 1/10. Then m = vu = -90-9 = 10 (taking v = -90 cm from 1/v = 1/f + 1/u with f = 10).

Step 2 — area of each square in the image. Area scales as (linear magnification)²: Area_image = |m|² × Area_object = 10² × 1 mm² = 100 mm² = 1 cm².

(b) Angular magnification (magnifying power). For a magnifying glass with the eye held close to the lens and the object at distance |u| from it, M = D|u| = 259 ≈ 2.78. This is the ratio of the angle subtended at the eye by the image (which equals the angle subtended by the object at distance |u|) to the angle subtended by the object held at the near point D = 25 cm.

(c) Are m and M equal? NO — they are different quantities. Here m = 10 (linear) while M ≈ 2.78 (angular). The two are equal only when the image is formed at the near point of the eye i.e., when |v| = D = 25 cm; here the image forms much farther away |v| = 90 cm, so the angular size at the eye is much smaller than the linear size at the image plane.

Final answer. m = 10, area = 100 mm², M = 25/9 ≈ 2.78. Not equal.

Concept used. Maximum magnifying power of a simple magnifier is obtained when the final image forms at the NEAR POINT D = -25 cm, with the negative sign for virtual image on the same side as object.

(a) Find u such that v = -25 cm, f = +9 cm. 1u = 1v - 1f = 1-25 - 19. LCM 225: 1u = -9225 - 25225 = -34225. So u = -22534 ≈ -6.62 cm. Hold the lens about 6.6 cm from the card.

(b) Magnification. m = vu = -25-6.62 = 256.62 ≈ 3.78.

(c) Magnifying power for near-point use. M = 1 + Df = 1 + 259 = 1 + 2.78 = 3.78. So magnification = magnifying power ≈ 3.78. They coincide because in this case the linear and angular ratios are both measured relative to the same reference (the near point).

Final answer. u ≈ -6.62 cm, m ≈ 3.78, M ≈ 3.78. Equal in this case.

Setup. Each square has actual area 1 mm², and we want the virtual image area 6.25 mm². Linear magnification (since area scales as linear)²): |m| = √image areaobject area = √6.251 = 2.5.

Step 1 — relate m to u, v. For a magnifying glass, m = v/u. Since the image is virtual, v < 0; the object is real, u < 0; so m = v/u > 0 (erect virtual image), and |m| = |v|/|u| = 2.5.

Step 2 — express v in terms of u. Take v = 2.5 u (both negative, so this preserves signs). Then apply lens formula 1/v - 1/u = 1/f: 12.5 u - 1u = 1f. 1u(12.5 - 1) = 1f, 1u·(1-2.52.5) = 1f. 1u×(-1.52.5) = 1f, u = -1.52.5 f = -0.6 f.

Step 3 — substitute f = 9 cm. u = -0.6 × 9 = -5.4 cm.

Step 4 — verify by finding v. v = 2.5 × (-5.4) = -13.5 cm. Check lens formula: 1v - 1u = 1-13.5 - 1-5.4 = -0.0741 + 0.185 = 0.1111 = 19.

Step 5 — can you see distinctly with eyes very close to the magnifier? The virtual image is at 13.5 cm from the lens, on the same side as the object. The near point of a normal eye is at 25 cm. Since |v| = 13.5 cm < 25 cm, the image is CLOSER than the near point — the eye cannot focus on it. So the image would appear blurred. NO, the squares cannot be seen distinctly with the eye close to the magnifier.

Final answer. u = -5.4 cm; NOT seen distinctly (image inside near point).

(a) Why a magnifying glass magnifies even though the angle is the "same." The angle subtended at the eye by the real object if held at distance D = 25 cm is ₀ = h/D. With the magnifier, the object can be placed much closer than D e.g., at ∼ f. Now the virtual image is at D (or infinity), but the linear angular size at the eye is the angle subtended by the object at its closer distance — which is much larger than ₀. So while the image subtends the same angle as the object would if held at the image location, that angle is much larger than the angle the real object would subtend if held at the near point.

(b) Does M change if the eye moves back? A little, but not much. When the eye is very close to the lens, it intercepts essentially all the rays the lens produces, and the angular size measured at the eye is maximal. Moving the eye back: the image's angular size at the eye decreases (since the image is at fixed distance from the lens, but now the eye is farther from both). The angular magnification decreases slightly. For practical purposes, the eye is held near the lens.

(c) Why not use ever-smaller f for ever-greater magnification? Three reasons:
(i) Aberrations. Short-focal-length lenses suffer severely from spherical and chromatic aberrations — distortions in the image. The image becomes blurred and coloured.
(ii) Manufacturing limits. A lens with f < 1 cm requires very high curvature, hard to grind precisely.
(iii) Useful magnification. The maximum useful magnification is set by the wavelength of light: features smaller than ∼ λ/2 ≈ 0.25 cannot be resolved no matter how strong the magnifier. Beyond this limit, you make the image bigger but reveal no more detail (called "empty magnification").

(d) Why both objective and eyepiece in a compound microscope must have short f. The total magnification is M = v_o/u_o× 1 + D/f_e. Short f_o means the object can be placed very close to the objective, giving large v_o/u_o. Short f_e makes the eyepiece's angular magnification 1+D/f_e large. Both factors multiply, so both lenses must be strong (short f) for high total magnification.

(e) Why the eye should be slightly back from the eyepiece. The eyepiece is designed so that all the light from the objective converges to a small region just behind the eyepiece — the "exit pupil" or "eye ring." If the eye is at this position, it captures all the light that emerged from the objective; if the eye is too close to the lens (in front of the exit pupil) or too far (beyond it), some light is lost, the image dims, and the field of view shrinks. The proper position is at the eye ring — typically 2–3 cm behind the eyepiece (the "eye relief"). Modern eyepieces are designed for eye relief comfortable for spectacle wearers.

Setup. Total magnifying power |M| = 30; image at near point D = 25 cm; f_o = 1.25 cm, f_e = 5 cm.

Step 1 — eyepiece angular magnification. M_e = 1 + Df_e = 1 + 255 = 6.

Step 2 — required objective magnification. |M_o| = |M|M_e = 306 = 5. So the objective must produce a 5-fold linear magnification of the object.

Step 3 — find u_o from M_o = v_o/u_o = -5 (real, inverted image). So v_o = -5 u_o. Combined with lens formula 1/v_o - 1/u_o = 1/f_o: 1-5 u_o - 1u_o = 11.25. -15 u_o - 1u_o = 11.25, -1+55 u_o = 11.25, -65 u_o = 11.25. u_o = -6× 1.255 = -1.5 cm.

Step 4 — find v_o. v_o = -5 u_o = -5 × (-1.5) = 7.5 cm.

Step 5 — find u_e (intermediate image is object for eyepiece, image at near point). 1-25 - 1u_e = 15, 1u_e = -125 - 15 = -125 - 525 = -625. u_e = -256 ≈ -4.17 cm.

Step 6 — lens separation. L = v_o + |u_e| = 7.5 + 4.17 = 11.67 cm ≈ 11.7 cm.

Final answer.

• Object placed at |u_o| = 1.5 cm in front of objective.
• Lens separation L ≈ 11.67 cm.

Given. f_o = 140 cm, f_e = 5.0 cm, D = 25 cm.

(a) Normal adjustment — final image at infinity. |M| = f_of_e = 1405 = 28.

(b) Final image at near point — magnification. |M| = f_of_e(1 + f_eD) = 28(1 + 525) = 28 × 1.2 = 33.6.

Final answer. (a) |M| = 28, (b) |M| = 33.6.

Given. f_o = 140 cm, f_e = 5.0 cm. Normal adjustment.

(a) Separation between objective and eyepiece in normal adjustment. L = f_o + f_e = 140 + 5 = 145 cm.

(b) Height of objective's image of a 100-m tower at 3 km.

Step 1 — angular size of tower. θ = 100 m3000 m = 130 rad.

Step 2 — linear height at focal plane. h_image = f_o · θ = 140 × 130 ≈ 4.67 cm.

(c) Final image height at near point D = 25 cm.

Step 1 — eyepiece magnification at near point. m_e = 1 + Df_e = 1 + 255 = 6.

Step 2 — final image height. h_final = m_e · h_image = 6 × 4.67 = 28 cm.

Final answer. (a) L = 145 cm, (b) h_int ≈ 4.67 cm, (c) h_final = 28 cm.

Setup. Cassegrain telescope: large primary concave mirror at the back, smaller secondary convex mirror near the front. Light enters through the front opening, reflects off the primary, then off the secondary (back through a hole in the primary).

Given. Primary radius R₁ = 220 mm concave, so f₁ = -110 mm. Secondary radius R₂ = 140 mm convex, so f₂ = +70 mm. Separation d = 20 mm.

Step 1 — primary mirror forms image at its focus. Light from infinity, primary forms an image at f₁ = 110 mm in front of itself (on the same side as the object). This intermediate image is the object for the secondary mirror.

Step 2 — locate intermediate image relative to secondary. Secondary is at distance d = 20 mm in front of the primary. The intermediate image is at 110 mm in front of primary, so it's 110 - 20 = 90 mm behind the secondary mirror. Behind the secondary = on the side away from where light is coming from = a virtual object. For the secondary (in standard Cartesian for mirrors, with light now travelling in the reverse direction after primary reflection): u₂ = +90 mm (virtual object, positive in the direction of light propagation after the first reflection? careful — let's set up signs).

Step 3 — apply mirror formula for secondary. Use f₂ = +70 mm (convex), u₂ = -90 mm (real object would be in front of mirror; here it's behind, which we treat by signing as positive but the formula needs care). NCERT solution: take u₂ = +90 mm for a virtual object behind the secondary mirror. 1v₂ + 1u₂ = 1f₂ ⇒ 1v₂ = 170 - 190. LCM 630: 1v₂ = 9630 - 7630 = 2630 = 1315. So v₂ = 315 mm.

Step 4 — locate final image relative to primary. The image is 315 mm in front of the secondary, hence 315 - 20 = 295 mm behind the primary (light has reflected back). So the final image forms 295 mm behind the primary mirror — exactly where the eyepiece is positioned in a real Cassegrain telescope.

Final answer. Final image is 315 mm in front of the secondary, or 295 mm behind the primary mirror.

Setup. A laser hits the mirror; when the mirror tilts by angle θ, the reflected beam rotates by 2θ. The spot on a distant screen moves by an amount determined by this angular shift and the distance.

Concept used — mirror tilt amplification. If a plane mirror rotates by angle θ, the reflected ray rotates by 2θ (the law of reflection guarantees the angle of incidence equals angle of reflection, so doubling).

Given. Mirror deflection θ = 3.5°, screen distance L = 1.5 m.

Step 1 — angular rotation of reflected beam. φ = 2θ = 2× 3.5° = 7.0°. Convert to radians: φ = 7° × π/180 = 0.1222 rad.

Step 2 — displacement of spot on screen. For small angles, Δ x = L tanφ ≈ L φ. Δ x = 1.5 × 0.1222 = 0.1833 m = 18.33 cm.

Step 3 — using tan for accuracy. tan 7° = 0.1228, so Δ x = 1.5 × 0.1228 = 0.1842 m = 18.42 cm. Either way, ≈ 18.4 cm.

Final answer. Δ x ≈ 18.4 cm.

Setup. An equiconvex glass lens sits on a plane mirror with a thin film of liquid between. When the needle (object) is moved along the axis until its image coincides with itself, the rays must hit the mirror at the back perpendicular to the mirror (so they retrace) — meaning the rays emerging from the lens-liquid combination must form a beam that converges toward the focal point of the equivalent lens-mirror system. Standard result: the needle is at the focal length f_eq of the equivalent system.

Step 1 — f_{lens alone} (from second measurement, no liquid). Without liquid, only the lens + mirror combination. The equivalent focal length f_lens of the lens-mirror system (lens then mirror, light retraces) is determined by: 1f_eq = 2f_lens + 1f_mirror. For a plane mirror, f_mirror = ∞, so 1/f_eq = 2/f_lens, i.e., f_eq = f_lens/2. But experimentally the focal length of the equivalent system is the needle distance = 30 cm. So f_lens/2 = 30, giving f_lens = 60 cm. Wait — the conventional analysis: needle at object position = |f_system|. The system "lens twice + mirror once" has power P_eq = 2P_lens + P_mirror. For plane mirror, P_mirror=0, so P_eq = 2 P_lens. Equivalent focal length f_eq = 1/2/f_lens = f_lens/2. With needle at 30 cm = f_eq: f_lens = 60 cm.

Step 2 — f_combination (with liquid). Now the lens system is: glass lens + plano-concave liquid lens (the liquid fills the space between the convex bottom of the glass lens and the flat mirror surface — so it's plano-concave with one curved face matching the glass lens's bottom, and one flat face matching the mirror). Combined power: P_comb = P_glass + P_liquid. Equivalent system: P_{eq, comb} = 2 P_comb + P_mirror = 2 P_comb. With needle at 45 cm: f_{eq, comb} = 45 cm, so P_{eq, comb} = 1/0.45. Then P_comb = 1/0.90, i.e., f_comb = 90 cm.

Step 3 — extract f_liquid. 1f_comb = 1f_glass + 1f_liquid ⇒ 190 = 160 + 1f_liquid. 1f_liquid = 190 - 160. LCM 180: 1f_liquid = 2180 - 3180 = -1180. So f_liquid = -180 cm (diverging, as expected for the plano-concave liquid lens).

Step 4 — find radius of curvature of glass lens surface. Equiconvex glass lens: R₁ = +R, R₂ = -R. Lensmaker: 1f_glass = (_g - 1)(1R - 1-R) = 2(_g - 1)R. With _g = 1.50 and f_glass = 60 cm: 160 = 2 × 0.5R = 1R ⇒ R = 60 cm.

Step 5 — find _l from the liquid lens. Liquid lens: plano-concave touching glass's bottom convex face on top with R = -60 cm, flat at the bottom with R = ∞. Wait — the liquid is BELOW the glass lens, so for the liquid lens taken in isolation: top surface has the same curvature as the glass lens's bottom surface which was R₂ = -60 cm for the glass lens, but now this is the "first surface" of the liquid lens, which makes its sign.... Best: use the convention that the liquid lens is plano-concave with R₁ = -60 cm (top, concave) and R₂ = ∞ (bottom, flat). 1f_liquid = (_l - 1)(1R₁ - 1R₂) = (_l - 1)(1-60 - 0) = -_l - 160. With f_liquid = -180 cm: -1180 = -_l - 160 ⇒ _l - 1 = 60180 = 13 ⇒ _l = 43 ≈ 1.33.

Final answer. _l ≈ 1.33 (water!).