Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
Try each problem first, then open the solution to check your working.
An object is placed 25 cm from a concave mirror of focal length 15 cm. Find the image distance and the magnification.
Given: f = −15 cm; u = −25 cm
Use: 1/v + 1/u = 1/f
1/v = 1/f − 1/u = −1/15 + 1/25 = −2/75
v = −37.5 cm (real, in front of the mirror)
m = −v/u = −37.5/25 = −1.5 (inverted, magnified)
Answer: 37.5 cm in front; m = −1.5
A convex mirror of radius 40 cm has an object 30 cm in front of it. Find the image position and magnification.
Given: R = 40 cm; f = +20 cm; u = −30 cm
Use: 1/v = 1/f − 1/u
1/v = 1/20 + 1/30 = 1/12
v = +12 cm (virtual, behind the mirror)
m = −v/u = 12/30 = 0.4 (erect, diminished)
Answer: 12 cm behind; m = 0.4
Where must an object be placed in front of a concave mirror of focal length 10 cm to get an image of the same size?
Given: f = −10 cm; m = −1
Use: For |m| = 1, v = u
m = −v/u = −1 gives v = u
1/u + 1/u = 1/f gives u = 2f = −20 cm
Answer: At the centre of curvature, 20 cm from the mirror
An object is 10 cm in front of a concave mirror of focal length 20 cm. Find the image and magnification.
Given: f = −20 cm; u = −10 cm
Use: 1/v = 1/f − 1/u
1/v = −1/20 + 1/10 = 1/20
v = +20 cm (virtual, behind the mirror)
m = −v/u = 20/10 = +2
Answer: 20 cm behind the mirror; m = +2
A shaving mirror of focal length 30 cm is to give an erect image 3 times the size of the face. How far should the face be?
Given: f = −30 cm; m = +3
Use: m = f/(f − u)
3 = −30/(−30 − u), so −30 − u = −10
u = −20 cm
Answer: 20 cm from the mirror
A concave mirror forms the image of the Sun at 12 cm from it. Find its radius of curvature.
Given: f = 12 cm (image of a distant object)
Use: R = 2f
R = 24 cm
Answer: 24 cm
A ray is incident at 45° on a surface and refracts at 30°. Find the refractive index.
Given: i = 45°; r = 30°
Use: n = sin i / sin r
n = 0.7071 / 0.5 = 1.414
Answer: 1.41
Find the speed of light in glass of refractive index 1.5.
Given: n = 1.5; c = 3 × 10⁸ m/s
Use: v = c/n
v = 3 × 10⁸ / 1.5 = 2 × 10⁸ m/s
Answer: 2 × 10⁸ m/s
A fish is 20 cm below the surface of water (n = 1.33). At what depth does it appear?
Given: d = 20 cm; n = 1.33
Use: d′ = d/n
d′ = 20/1.33 = 15.0 cm
Answer: 15 cm
Find the critical angle for glass (n = 1.5) and for water (n = 1.33) in air.
Given: n = 1.5 and 1.33
Use: sin θ_c = 1/n
Glass: sin θ_c = 0.667, θ_c = 41.8°
Water: sin θ_c = 0.752, θ_c = 48.8°
Answer: 41.8°; 48.8°
A 6 cm thick glass slab (n = 1.5) lies on a printed page. By how much is the print raised when viewed from above?
Given: t = 6 cm; n = 1.5
Use: shift = t (1 − 1/n)
shift = 6 × (1 − 0.667) = 2 cm
Answer: 2 cm
What is the least refractive index of a right-angled isosceles prism so that a ray entering normally is totally reflected at the hypotenuse?
Given: Angle of incidence at the hypotenuse = 45°
Use: Need sin 45° ≥ 1/n
n ≥ 1/sin 45° = √2 = 1.41
Answer: n ≥ 1.41
A point source is 40 cm below the surface of water (n = 1.33). Find the radius of the circle of the surface through which light emerges.
Given: h = 40 cm; n = 1.33
Use: r = h / √(n² − 1)
r = 40 / √(1.769 − 1) = 40 / 0.877 = 45.6 cm
Answer: 45.6 cm
A point object is 20 cm from a convex refracting surface of radius 10 cm separating air from glass (n = 1.5). Where is the image?
Given: n₁ = 1; n₂ = 1.5; R = +10 cm; u = −20 cm
Use: n₂/v − n₁/u = (n₂ − n₁)/R
1.5/v + 1/20 = 0.5/10 = 0.05
1.5/v = 0, so v = ∞: the refracted rays are parallel
Answer: At infinity (the object is at the first focus)
A thin lens of glass (n = 1.5) has R₁ = +20 cm and R₂ = −20 cm. Find its focal length.
Given: n = 1.5; R₁ = 20 cm; R₂ = −20 cm
Use: 1/f = (n − 1)(1/R₁ − 1/R₂)
1/f = 0.5 (1/20 + 1/20) = 0.05
f = 20 cm
Answer: +20 cm
The lens above (f = 20 cm in air) is placed in water (n = 1.33). Find its focal length.
Given: n_g = 1.5; n_w = 1.33
Use: 1/f′ = (n_g/n_w − 1)(1/R₁ − 1/R₂)
1/f′ = (1.128 − 1)(0.1) = 0.01278
f′ = 78 cm
Answer: 78 cm (longer than in air)
A convex lens of focal length 10 cm has an object 15 cm in front of it. Find the image and the magnification.
Given: f = +10 cm; u = −15 cm
Use: 1/v = 1/f + 1/u
1/v = 1/10 − 1/15 = 1/30
v = +30 cm (real)
m = v/u = 30/(−15) = −2 (inverted, magnified)
Answer: 30 cm on the other side; m = −2
A concave lens of focal length 20 cm has an object 30 cm in front of it. Find the image and magnification.
Given: f = −20 cm; u = −30 cm
Use: 1/v = 1/f + 1/u
1/v = −1/20 − 1/30 = −1/12
v = −12 cm (virtual)
m = v/u = 0.4
Answer: 12 cm on the same side; m = 0.4
A lens of +5 D is in contact with a lens of −2 D. Find the power and focal length of the combination.
Given: P₁ = +5 D; P₂ = −2 D
Use: P = P₁ + P₂
P = +3 D
f = 1/3 m = 33.3 cm (converging)
Answer: +3 D; 33 cm
A 5 cm tall object is 30 cm from a convex lens of focal length 20 cm. Find the image height.
Given: f = 20 cm; u = −30 cm; h = 5 cm
Use: 1/v = 1/20 − 1/30
v = +60 cm
m = v/u = −2
image height = 10 cm (inverted)
Answer: 10 cm, real and inverted
A convex lens of f = 20 cm is in contact with a concave lens of f = 30 cm. Find the focal length of the combination.
Given: f₁ = +20 cm; f₂ = −30 cm
Use: 1/f = 1/f₁ + 1/f₂
1/f = 1/20 − 1/30 = 1/60
f = +60 cm
Answer: +60 cm (converging)
A prism of angle 60° has n = 1.5. Find the angle of minimum deviation.
Given: A = 60°; n = 1.5
Use: sin[(A + D)/2] = n sin(A/2)
sin[(60° + D)/2] = 1.5 × 0.5 = 0.75
(60° + D)/2 = 48.6°, so D = 37.2°
Answer: 37.2°
A thin prism of angle 4° has n = 1.5. Find the deviation.
Given: A = 4°; n = 1.5
Use: δ = (n − 1) A
δ = 0.5 × 4° = 2°
Answer: 2°
A prism of angle 60° gives a minimum deviation of 30°. Find n.
Given: A = 60°; D = 30°
Use: n = sin[(A + D)/2] / sin(A/2)
n = sin 45° / sin 30° = 0.7071/0.5 = 1.414
Answer: 1.41
A ray is refracted in a prism at r₁ = 30° and leaves at r₂ = 25° inside. Find the angle of the prism.
Given: r₁ = 30°; r₂ = 25°
Use: A = r₁ + r₂
A = 55°
Answer: 55°
A convex lens of focal length 5 cm is used as a magnifier. Find the angular magnification with the image at (a) the near point 25 cm and (b) infinity.
Given: f = 5 cm; D = 25 cm
Use: M = 1 + D/f; M = D/f
(a) M = 1 + 25/5 = 6
(b) M = 25/5 = 5
Answer: 6; 5
A compound microscope has f_o = 1 cm and f_e = 5 cm, with a tube length (distance between the lenses) of about 15 cm. Estimate its magnifying power.
Given: f_o = 1 cm; f_e = 5 cm; L ≈ 15 cm
Use: M ≈ (L/f_o)(D/f_e)
M ≈ (15/1)(25/5) = 75
Answer: about 75
A telescope has f_o = 100 cm and f_e = 4 cm. Find its magnifying power and length in normal adjustment.
Given: f_o = 100 cm; f_e = 4 cm
Use: M = f_o/f_e; L = f_o + f_e
M = 100/4 = 25
L = 104 cm
Answer: 25; 104 cm
A person has a near point of 50 cm. What lens would let him read at 25 cm?
Given: Near point 50 cm; reading distance 25 cm
Use: 1/f = 1/v − 1/u with u = −25, v = −50
1/f = −1/50 + 1/25 = 1/50 cm⁻¹
f = +50 cm, P = +2 D (a convex lens)
Answer: +2 D
A person has a far point at 2 m. What lens lets him see distant objects?
Given: Far point 2 m
Use: The lens must form an image of infinity at the far point
f = −2 m (concave)
P = −0.5 D
Answer: −0.5 D
The objective of a microscope has f = 0.5 cm and the object is 0.55 cm from it. Find the image distance and the magnification.
Given: f_o = 0.5 cm; u = −0.55 cm
Use: 1/v = 1/f + 1/u
1/v = 2 − 1.818 = 0.1818, so v = 5.5 cm
m_o = v/u = −10
Answer: 5.5 cm; m = −10
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
f = −R/2 = −18 cm, u = −27 cm. 1/v = 1/f − 1/u = −1/18 + 1/27 = −1/54, so v = −54 cm: the screen should be 54 cm in front of the mirror.
m = −v/u = −54/27 = −2. The image is real, inverted and magnified; its size is 2 × 2.5 = 5.0 cm.
If the candle is moved closer to the mirror (towards the focus), the image moves farther away and becomes larger, so the screen must be moved away from the mirror.
f = +15 cm, u = −12 cm. 1/v = 1/f − 1/u = 1/15 + 1/12 = 9/60, so v = +6.7 cm: a virtual image 6.7 cm behind the mirror.
m = −v/u = 6.67/12 = 0.56. The image is erect and 4.5 × 0.56 = 2.5 cm tall.
As the needle moves away, the image moves towards the focus (it approaches v = 15 cm but never reaches it) and becomes smaller (diminished), always virtual and erect.
n = real depth / apparent depth = 12.5 / 9.4 = 1.33.
In the liquid of n = 1.63 the apparent depth is 12.5 / 1.63 = 7.67 cm. The needle now appears 9.4 − 7.67 = 1.73 cm nearer the top, so the microscope must be moved up by 1.7 cm.
From Fig. (a): n_g = sin 60° / sin 35° = 0.866 / 0.574 = 1.51. From Fig. (b): n_w = sin 60° / sin 47° = 0.866 / 0.731 = 1.18.
At the water–glass surface: n_w sin 45° = n_g sin r, so sin r = (1.18 × 0.707) / 1.51 = 0.553 and r = 33.6° ≈ 34°.
Light emerges only for angles of incidence less than the critical angle: sin θ_c = 1/1.33 = 0.752, θ_c = 48.75°, tan θ_c = 1.139.
Radius of the circle: r = h tan θ_c = 80 × 1.139 = 91.2 cm = 0.912 m.
Area = π r² = π (0.912)² = 2.6 m².
n = sin[(A + D)/2] / sin(A/2) = sin 50° / sin 30° = 0.766 / 0.5 = 1.53.
In water the relative refractive index is 1.532 / 1.33 = 1.152. Then sin[(60° + D′)/2] = 1.152 × sin 30° = 0.576, so (60° + D′)/2 = 35.2° and D′ ≈ 10.3°.
1/f = (n − 1)(1/R₁ − 1/R₂) with R₁ = R and R₂ = −R: 1/f = (n − 1) 2/R, so R = 2 (n − 1) f = 2 × 0.55 × 20 = 22 cm.
The point P acts as a virtual object for the lens, at u = +12 cm (on the side where the light emerges).
(a) Convex, f = +20 cm: 1/v = 1/f + 1/u = 1/20 + 1/12 = 8/60, so v = 7.5 cm from the lens (on the far side): the beam converges 4.5 cm before P.
(b) Concave, f = −16 cm: 1/v = −1/16 + 1/12 = 1/48, so v = 48 cm from the lens: the beam converges 36 cm beyond P.
f = −21 cm, u = −14 cm: 1/v = 1/f + 1/u = −1/21 − 1/14 = −5/42, so v = −8.4 cm: a virtual image, 8.4 cm from the lens on the same side as the object.
m = v/u = 8.4/14 = 0.6, so the image is erect and diminished, of height 3.0 × 0.6 = 1.8 cm.
As the object is moved farther away, the virtual image moves towards the focus (always inside f = 21 cm) and becomes smaller, remaining erect and virtual.
1/f = 1/f₁ + 1/f₂ = 1/30 − 1/20 = −1/60, so f = −60 cm. The combination is a diverging lens (negative focal length).
(a) Final image at 25 cm: for the eyepiece v_e = −25 cm, f_e = 6.25 cm: 1/u_e = 1/v_e − 1/f_e = −1/25 − 4/25 = −1/5, so u_e = −5 cm. The objective's image is therefore 5 cm from the eyepiece, v_o = 15 − 5 = 10 cm. For the objective: 1/u_o = 1/v_o − 1/f_o = 1/10 − 1/2 = −2/5, so u_o = −2.5 cm (the object is 2.5 cm from the objective).
m_o = v_o/u_o = 10/(−2.5) = −4 and m_e = 1 + D/f_e = 1 + 25/6.25 = 5. Magnifying power M = m_o m_e = −20 (magnitude 20).
(b) Final image at infinity: u_e = −f_e = −6.25 cm, so v_o = 15 − 6.25 = 8.75 cm. 1/u_o = 1/8.75 − 1/2 gives u_o = −2.59 cm.
m_o = 8.75/(−2.59) = −3.38 and m_e = D/f_e = 25/6.25 = 4. M = −13.5 (magnitude 13.5).
Objective: f_o = 0.8 cm, u_o = −0.9 cm: 1/v_o = 1/0.8 − 1/0.9 = 0.1389, so v_o = 7.2 cm.
Eyepiece: the final image is at the near point, v_e = −25 cm: 1/u_e = 1/v_e − 1/f_e = −1/25 − 1/2.5 = −11/25, so u_e = −2.27 cm.
Separation = v_o + |u_e| = 7.2 + 2.27 = 9.47 cm.
M = m_o m_e = (7.2/0.9) × (1 + 25/2.5) = 8 × 11 = 88.
In normal adjustment M = f_o / f_e = 144 / 6.0 = 24. The separation is f_o + f_e = 144 + 6.0 = 150 cm.
(a) M = f_o / f_e = 15 / 0.01 = 1500.
(b) The angle subtended by the Moon at the objective: θ = 3.48 × 10⁶ / 3.8 × 10⁸ = 9.16 × 10⁻³ rad. The image formed by the objective lies at its focus, so its diameter = f_o θ = 15 × 9.16 × 10⁻³ = 0.137 m ≈ 13.7 cm.
Use 1/v = 1/f − 1/u (Cartesian convention: for a real object u < 0).
(a) Concave mirror, f < 0, and −2|f| < u < −|f|, i.e. |f| < |u| < 2|f|. Then 1/|v| = 1/|u| − 1/|f| in magnitude form: 1/|v| = 1/|f| − 1/|u| < 1/|f| − 1/(2|f|) = 1/(2|f|), so |v| > 2|f| and v is negative: the image is real and beyond 2f.
(b) Convex mirror, f > 0 and u < 0: 1/v = 1/f − 1/u = 1/f + 1/|u| > 0, so v is always positive (behind the mirror): the image is always virtual.
(c) With v positive, 1/v = 1/f + 1/|u| > 1/f, so v < f: the image lies between the pole and the focus. Also m = −v/u = v/|u| = f/(f + |u|) < 1: the image is diminished.
(d) Concave mirror, |u| < |f|: 1/v = 1/f − 1/u = −1/|f| + 1/|u| > 0, so v is positive (virtual image behind the mirror). m = −v/u = f/(f − u) = −|f| / (−|f| + |u|) = |f|/(|f| − |u|) > 1: the image is enlarged and erect.
The slab makes the pin appear raised by t (1 − 1/n) = 15 × (1 − 1/1.5) = 15 × 1/3 = 5 cm.
The answer does not depend on where the slab is placed between the eye and the pin (for rays close to the normal).
(a) Total reflection at the core–cladding boundary needs sin θ_c = 1.44/1.68 = 0.857, θ_c = 59°. So the ray must make an angle of at least 59° with the normal to the wall, i.e. at most 90° − 59° = 31° with the axis inside the pipe (r ≤ 31°).
At the end face, sin i = n sin r = 1.68 × sin 31° = 0.865, so i = 60°. Total reflection takes place for rays with 0 < i < 60° to the axis.
(b) With no covering the boundary is glass–air: sin θ_c = 1/1.68, θ_c = 36.5°, so r ≤ 53.5° would be needed. Then sin i = 1.68 × sin 53.5° = 1.35 > 1, which is impossible: for every angle of incidence from 0° to 90° the ray undergoes total internal reflection inside the pipe.
A real image needs an object–image distance D ≥ 4f. So f ≤ D/4 = 3/4 = 0.75 m. The maximum focal length is 75 cm (at exactly this value the image forms with u = v = 1.5 m).
For the displacement method f = (D² − d²) / 4D, with D = 90 cm and d = 20 cm.
f = (8100 − 400) / 360 = 7700 / 360 = 21.4 cm.
(a) f₁ = +30 cm, f₂ = −20 cm, d = 8 cm: 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂) = 1/30 − 1/20 + 8/600 = (20 − 30 + 8)/600 = −2/600. So F = −300 cm.
The positions at which F is measured (the principal planes) are different for the two sides, so the effective focal length depends on the side from which the parallel beam arrives (the distance of the focus from the lens combination differs). With the lenses separated, the notion of a single effective focal length is of limited use: it is better to find the image by applying the lens formula to each lens in turn, as in part (b).
(b) Convex lens, u = −40 cm, f = +30 cm: 1/v₁ = 1/30 − 1/40 = 1/120, v₁ = 120 cm; m₁ = v₁/u = −3. This image is 120 − 8 = 112 cm from the concave lens, so u₂ = −112 cm for it. For the concave lens (f = −20 cm): 1/v₂ = −1/20 − 1/112 = −132/2240, v₂ = −16.97 cm; m₂ = v₂/u₂ = 0.152.
Net magnification m = m₁ m₂ = −3 × 0.152 = −0.45. Image size = 0.45 × 1.5 = 0.68 cm (inverted).
At the second face the angle of incidence must equal the critical angle: sin r₂ = 1/1.524 = 0.656, so r₂ = 41.0°.
r₁ + r₂ = A gives r₁ = 60° − 41.0° = 19.0°.
Snell's law at the first face: sin i = n sin r₁ = 1.524 × sin 19.0° = 0.496, so i ≈ 30°.
Note on the data: with f = 9 cm and the card at 9 cm, the card lies at the focus and the image is at infinity, so the linear magnification is not finite. NCERT's answer key was worked with f = 10 cm; the answers below use f = 10 cm (u = −9 cm) so that they match it.
(a) 1/v = 1/f + 1/u = 1/10 − 1/9 = −1/90, so v = −90 cm (virtual image 90 cm from the lens). m = v/u = 90/9 = 10. The area of each square in the image = 1 mm² × 10² = 100 mm² = 1 cm².
(b) The magnifying power is M = D/|u| = 25/9 = 2.8 (the ratio of the angle subtended by the image to that subtended by the card at the near point, 25 cm).
(c) No. The magnification (10) is the ratio of the sizes of the image and object, while the magnifying power (2.8) compares angles. They are equal only when the image is formed at the near point, 25 cm. Here the image is at 90 cm.
(a) The maximum magnifying power is obtained with the image at the near point, v = −25 cm. With f = 10 cm: 1/u = 1/v − 1/f = −1/25 − 1/10 = −7/50, so u = −7.14 cm: hold the lens 7.14 cm from the card.
(b) m = v/u = 25/7.14 = 3.5.
(c) Yes. The magnifying power M = 1 + D/f = 1 + 25/10 = 3.5 is equal to the magnification since the image is at the near point (the eye is close to the lens, so the image is 25 cm from the eye).
The area is 6.25 times the original, so the linear magnification is m = √6.25 = 2.5. With m = v/u = 2.5, v = 2.5u (both negative). Lens formula with f = 10 cm: 1/(2.5u) − 1/u = 1/10, so −1.5/(2.5u) = 1/10 and u = −6 cm.
Then v = −15 cm. The virtual image is only 15 cm from the lens (and the eye), which is less than the near point 25 cm, so you would not be able to see the squares distinctly.
(a) Without the lens the object cannot be brought closer to the eye than the near point (25 cm) and still be seen clearly, so the largest angle it can subtend is that at 25 cm. With the magnifier the object is placed close to the lens (within f), and the virtual image, which can still be viewed at 25 cm or more, subtends a larger angle than the object would at 25 cm. The angular magnification is the ratio of these two angles (it is relative to the unaided eye at the near point).
(b) Yes, it decreases slightly. Moving the eye back from the lens makes the virtual image subtend a smaller angle at the eye, and some of the rays refracted by the lens do not enter the eye (the field of view also shrinks).
(c) Lenses of very short focal length have very small radii of curvature and are hard to make and to handle. They also suffer from large spherical and chromatic aberrations, and the fields of view become very small. So in practice a simple microscope has a magnifying power of at most about 10 (very short lenses can reach 30 or so with great difficulty).
(d) For a compound microscope M ≈ (L/f_o)(D/f_e). For a large M both f_o and f_e must be small (for the given L); a short f_o gives a large magnification by the objective and a short f_e a large magnification by the eyepiece. Also, since the object must be placed just beyond f_o and the intermediate image just inside f_e, the lenses need short focal lengths to keep the instrument compact.
(e) The rays from the objective, refracted by the eyepiece, form a small bright circle called the eye-ring (the image of the objective formed by the eyepiece). All the light passing through the microscope passes through this circle, so the eye should be placed there to receive the maximum light and field of view. Its distance from the eyepiece is the image distance for the objective as the object: for the microscope of Exercise 9.12 (eyepiece f = 2.5 cm, objective 9.47 cm from it) it is v = 1/(1/2.5 − 1/9.47) = 3.4 cm; in general about 2–3 cm.
Take the final image at the near point (25 cm). Eyepiece: m_e = 1 + D/f_e = 1 + 25/5 = 6. So the objective must have m_o = 30/6 = 5, i.e. v_o = 5|u_o|.
Objective: 1/v_o − 1/u_o = 1/f_o with u_o = −x, v_o = 5x: 1/(5x) + 1/x = 1/1.25, so 1.2/x = 0.8 and x = 1.5 cm. So u_o = −1.5 cm and v_o = 7.5 cm.
Eyepiece: v_e = −25 cm, 1/u_e = −1/25 − 1/5 = −6/25, so u_e = −4.17 cm.
Separation between the lenses = v_o + |u_e| = 7.5 + 4.17 = 11.67 cm. So place the object 1.5 cm from the objective and keep the lenses 11.67 cm apart.
(a) M = f_o / f_e = 140 / 5.0 = 28.
(b) M = (f_o/f_e)(1 + f_e/D) = 28 × (1 + 5/25) = 28 × 1.2 = 33.6.
(a) Separation = f_o + f_e = 140 + 5.0 = 145 cm.
(b) The angle subtended by the tower = 100 / 3000 = 1/30 rad. The objective forms the image at its focus, of height h = f_o × (1/30) = 140/30 = 4.7 cm.
(c) The eyepiece, with the final image at 25 cm, has magnification 1 + D/f_e = 1 + 25/5 = 6. The height of the final image = 4.67 × 6 = 28 cm.
The large (primary, concave) mirror has f₁ = 220/2 = 110 mm. For an object at infinity it would form the image at 110 mm from it. The small (secondary, convex) mirror is 20 mm from the primary, so this image lies 110 − 20 = 90 mm behind the secondary mirror: it acts as a virtual object for the secondary, with u = +90 mm.
The secondary is convex with f₂ = +140/2 = +70 mm. Mirror formula: 1/v = 1/f₂ − 1/u = 1/70 − 1/90 = 2/630, so v = 315 mm.
The final image is real, 315 mm from the secondary mirror, on its reflecting side (towards the primary), so it forms behind the primary mirror, which has a central hole for the eyepiece.
When a mirror turns through an angle θ, the reflected ray turns through 2θ = 7.0°.
The displacement of the spot on the screen d = 1.5 × tan 7.0° = 1.5 × 0.1228 = 0.184 m = 18.4 cm.
The image coincides with the needle when the rays fall normally on the plane mirror and retrace their path; this happens when the needle is at the focal point of the lens (system) so that the rays emerge parallel.
Without the liquid the distance 30.0 cm is the focal length of the equiconvex lens: f₁ = 30 cm. For an equiconvex lens 1/f₁ = (n − 1)(2/R) = 0.5 × 2/R, so R = 30 cm.
With the liquid the focal length of the combination is f = 45 cm. The liquid forms a plano-concave lens with focal length f₂: 1/f = 1/f₁ + 1/f₂, so 1/f₂ = 1/45 − 1/30 = −1/90, f₂ = −90 cm.
For the plano-concave liquid lens (R₁ = −30 cm, R₂ = ∞): 1/f₂ = (n_l − 1)(1/R₁ − 1/R₂) = (n_l − 1)(−1/30). So −1/90 = −(n_l − 1)/30, giving n_l − 1 = 1/3 and n_l = 1.33.