Chapter 9: Ray Optics and Optical Instruments
Key Formulas:
• Mirror: 1/v + 1/u = 1/f
• Lens: 1/v − 1/u = 1/f (thin lens formula)
• Snell's Law: n₁ sinθ₁ = n₂ sinθ₂
• Prism: n = sin[(A + δm)/2] / sin(A/2)
Q1. An object 5 cm tall is placed 30 cm from a convex mirror of focal length 15 cm. Find the position and size of the image.
Solution
1/v + 1/u = 1/f
1/v + 1/(−30) = 1/15 (convex mirror, f = +15)
1/v = 1/15 + 1/30 = 3/30 = 1/10
v = 10 cm (behind the mirror)
m = −v/u = −10/(−30) = 1/3
Image height = 5 × 1/3 = 5/3 cm ≈ 1.67 cm (virtual, erect)
Q2. A candle is placed 20 cm from a convex lens of focal length 10 cm. At what distance from the lens should a screen be placed to get a sharp image? What is the nature of the image?
Solution
1/v − 1/u = 1/f
1/v − 1/(−20) = 1/10
1/v = 1/10 − 1/20 = 1/20
v = 20 cm (on the other side)
m = v/u = 20/(−20) = −1
Image is real, inverted, same size
Q3. A ray of light passes from glass (n = 1.5) to air. Find the critical angle.
Solution
sinθc = 1/n = 1/1.5 = 2/3
θc = sin−1(2/3) = 41.8°
Q4. Two thin lenses of focal lengths 20 cm and −30 cm are placed in contact. Find the equivalent focal length.
Solution
1/F = 1/f₁ + 1/f₂ = 1/20 + 1/(−30)
= 1/20 − 1/30 = (3−2)/60 = 1/60
F = 60 cm (converging combination)
Q5. A prism has angle A = 60° and minimum deviation δm = 37°. Find the refractive index of the material of the prism.
Solution
n = sin[(A + δm)/2] / sin(A/2)
= sin[(60 + 37)/2] / sin(30)
= sin(48.5°) / 0.5
= 0.749/0.5
= 1.5
Q6. Draw a ray diagram for a simple microscope when the object is at the near point. Derive the expression for magnifying power.
Solution
When object is at near point (D = 25 cm):
1/v − 1/u = 1/f
1/v = 1/f + 1/(−D) = 1/f − 1/D
Angular magnification: m = D/|u|
From lens formula: 1/|u| = 1/f − 1/D = (D−f)/(fD)
|u| = fD/(D−f)
m = D/|u| = (D−f)/f = D/f − 1 = 1 + D/f
(using D = 25 cm: m = 1 + 25/f)
Q7. An object is placed 20 cm from a concave mirror of focal length 10 cm. Find (i) the image position (ii) the magnification (iii) the nature of the image.
Solution
1/v + 1/u = 1/f
1/v + 1/(−20) = 1/(−10) (concave mirror, u and f negative)
1/v = −1/10 + 1/20 = (−2+1)/20 = −1/20
v = −20 cm (in front of the mirror)
m = −v/u = −(−20)/(−20) = −1
Image is real, inverted, same size (at 20 cm in front).
Q8. A glass slab of thickness 4 cm and refractive index 1.5 is placed in front of a mirror. An object is placed behind the slab at a distance of 8 cm from it. Find the apparent distance of the object from the mirror.
Solution
Apparent shift = t(1 − 1/n)
= 4(1 − 1/1.5) = 4(1/3) = 4/3 cm
Apparent distance = 8 − 4/3 = 6.67 cm from the slab
Distance from mirror = 4 + 6.67 = 10.67 cm
Q9. What is total internal reflection? Prove that the critical angle for total internal reflection is given by sin C = n₂/n₁.
Solution
Total internal reflection (TIR): When light travels from a denser to a rarer medium, beyond the critical angle, the light is completely reflected back into the denser medium.
Proof: Using Snell's law at the critical angle C, the refracted ray grazes the surface (θ₂ = 90°):
n₁ sin C = n₂ sin 90°
sin C = n₂/n₁
Q10. An optical fiber has a core of refractive index 1.5 and cladding of refractive index 1.4. Find the numerical aperture and the maximum angle of acceptance.
Solution
NA = √(n₁² − n₂²)
= √(1.5² − 1.4²)
= √(2.25 − 1.96) = √0.29
= 0.539
Acceptance angle: sinθa = NA = 0.539
θa = sin−1(0.539) = 32.6°
Q11. A compound microscope has an objective of focal length 2 cm and an eyepiece of focal length 5 cm. If the final image is formed at the near point (25 cm), find the magnifying power when the object is at 2.4 cm from the objective.
Solution
For objective: 1/v₁ − 1/u₁ = 1/f₁
1/v₁ = 1/2 + 1/(−2.4) = 0.5 − 0.4167 = 0.0833
v₁ = 12 cm
m₁ = v₁/u₁ = 12/(−2.4) = −5
For eyepiece (at near point):
m₂ = 1 + D/f₂ = 1 + 25/5 = 6
Total magnification: m = m₁ × m₂ = (−5)(6) = −30 (inverted)
Q12. An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. Find (i) the magnifying power (ii) the length of the telescope for normal adjustment.
Solution
(i) For normal adjustment:
M = −f₁/f₂ = −100/5 = −20
(ii) Length: L = f₁ + f₂ = 100 + 5 = 105 cm
Q13. A ray of light is incident at 60° on a glass slab of thickness 2 cm and refractive index √3. Find the lateral displacement of the ray.
Solution
Using Snell's law: n₁ sin i = n₂ sin r
sin r = sin 60°/√3 = (√3/2)/√3 = 1/2
r = 30°
Lateral displacement: d = t sin(i − r)/cos r
= 2 × sin(60° − 30°)/cos 30°
= 2 × sin 30°/cos 30° = 2 × tan 30°
= 2 × (1/√3) = 1.15 cm
Q14. A convex lens of focal length 20 cm forms a real image of an object placed 30 cm from it. Find the image position and magnification. What happens if a concave lens of focal length 10 cm is placed in contact with it?
Solution
1/v − 1/u = 1/f
1/v = 1/20 + 1/(−30) = (3−2)/60 = 1/60
v = 60 cm
m = v/u = 60/(−30) = −2 (real, inverted, enlarged)
Combined focal length:
1/F = 1/20 + 1/(−10) = 1/20 − 2/20 = −1/20
F = −20 cm (diverging combination)
Q15. The refractive index of a prism is √2 and its angle is 30°. Find the angle of minimum deviation and the angle of emergence when light is incident at the angle of minimum deviation.
Solution
n = sin[(A + δm)/2] / sin(A/2)
√2 = sin[(30 + δm)/2] / sin 15°
sin[(30 + δm)/2] = √2 × 0.2588 = 0.366
(30 + δm)/2 = sin−1(0.366) = 21.5°
δm = 13°
At minimum deviation, angle of emergence = angle of incidence = (A + δm)/2 = 21.5°
Q16. An air bubble is inside water of refractive index 4/3. Find the apparent depth of the bubble if it is at a real depth of 12 cm.
Solution
Apparent depth = Real depth / n
= 12/(4/3) = 12 × 3/4 = 9 cm
Q17. A diver is at a depth of 12 m in water. The refractive index of water is 4/3. Find (i) the apparent depth (ii) the apparent height of the sky as seen by the diver.
Solution
(i) Apparent depth = 12/(4/3) = 9 m
(ii) Apparent height of sky (snell's window):
The apparent height is determined by the critical angle:
sin C = 1/n = 3/4
The sky appears as a circle of angle 2C above the diver.
Apparent height of objects above water (from inside):
h' = nh = (4/3) × apparent...
The sky appears compressed into a cone of semi-angle C = sin−1(3/4) = 48.6°
Q18. Draw the ray diagram for an object placed between F and 2F of a convex lens. State the nature and position of the image formed.
Solution
When F < u < 2F:
• Image is formed beyond 2F on the other side
• Image is real, inverted, and enlarged
• 2F < v < ∞
Ray diagram: A ray parallel to the principal axis passes through F on the other side. A ray through the optical centre goes undeviated. They meet beyond 2F.
Q19. Two thin lenses of focal lengths +15 cm and −10 cm are separated by 5 cm along the principal axis. Find the equivalent focal length and the position of the principal planes.
Solution
1/F = 1/f₁ + 1/f₂ − d/(f₁f₂)
= 1/15 + 1/(−10) − 5/(15 × (−10))
= 1/15 − 1/10 + 5/150
= (10 − 15 + 5)/150 = 0/150
F = ∞ (afocal system — acts as a telescope-like combination)
Q20. A light ray incidents at 45° on one face of a prism of refracting angle 60°. The emergent ray makes an angle of 30° with the second face. Find the refractive index of the prism.
Solution
Using prism formula: r₁ + r₂ = A
From Snell's law at first face:
sin 45° = n sin r₁ ⇒ sin r₁ = 0.707/n
At second face: n sin r₂ = sin e
r₁ + r₂ = 60°
With e = 30°: n sin r₂ = sin 30° = 0.5
r₂ = sin−1(0.5/n)
r₁ = sin−1(0.707/n)
Solving: n ≈ 1.5
Q21. What is a reflecting telescope? Draw its labelled diagram and explain its advantages over a refracting telescope.
Solution
A reflecting telescope uses a concave mirror (objective) instead of a lens to gather light.
Advantages over refracting telescope:
• No chromatic aberration (mirrors don't disperse light)
• No spherical aberration (with parabolic mirrors)
• Light is reflected, not absorbed — more light collected
• Larger apertures possible (no sagging under weight)
• Cheaper to build for same aperture
Types: Newtonian, Cassegrain