Ray Optics and Optical Instruments — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Ray Optics and Optical Instruments" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Ray Optics and Optical Instruments" — 8 important questions with detailed answers for…
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NCERT Solutions for Class 12 Physics — Ray Optics and Optical Instruments. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Ray Optics and Optical Instruments — All 8 Questions Solved
Q1. What is refraction? State Snell's law and derive it using Huygens' principle.
Refraction: Bending of light ray at the interface between two media of different optical densities.
Snell's Law: n₁ sin θ₁ = n₂ sin θ₂
Where n₁, n₂ are refractive indices; θ₁, θ₂ are angles of incidence and refraction.
Derivation using Huygens' Principle:
Consider plane wavefronts AB and A'B' incident on interface at angle θ₁.
Let speed of light: v₁ in medium 1, v₂ in medium 2
Time taken from B to B': t = BB'/v₁ = AB'/v₂
In triangle ABB':
BB' = AB' sin θ₁
In triangle AA'B':
AA' = AB' sin θ₂
From BB'/v₁ = AA'/v₂:
AB' sin θ₁/v₁ = AB' sin θ₂/v₂
sin θ₁/v₁ = sin θ₂/v₂
Since n = c/v:
sin θ₁/(c/n₁) = sin θ₂/(c/n₂)
n₁ sin θ₁ = n₂ sin θ₂
This is Snell's law of refraction.
Q2. Define critical angle and total internal reflection. For glass (n = 1.5), find critical angle.
Critical Angle: Angle of incidence in denser medium for which angle of refraction is 90°.
Total Internal Reflection: Complete reflection of light at interface from denser to rarer medium when angle of incidence exceeds critical angle.
Derivation:
At critical angle θ_c, refraction angle = 90°
Using Snell's law: n₁ sin θ_c = n₂ sin 90°
n₁ sin θ_c = n₂ × 1
sin θ_c = n₂/n₁ = 1/n (for denser to rarer medium)
θ_c = sin⁻¹(1/n)
For glass-air interface (n = 1.5):
sin θ_c = 1/1.5 = 2/3
θ_c = sin⁻¹(0.667) ≈ 41.8°
Conditions for TIR:
1. Light travels from denser to rarer medium
2. Angle of incidence ≥ critical angle
3. No refracted ray is produced
4. Angle of reflection = angle of incidence
5. Both incident and reflected rays obey laws of reflection
Applications: Optical fibers, prisms (45-45-90 prism), diamonds
Q3. Derive lens maker's formula and show how it relates object distance, image distance, and focal length.
Lens Maker's Formula: 1/f = (n - 1)[1/R₁ - 1/R₂]
Where n = refractive index, R₁, R₂ = radii of curvature
Derivation:
Consider a thin lens with radii R₁ (first surface) and R₂ (second surface).
For refraction at first surface (air to glass):
1/s₁ + n/s₁' = (n-1)/R₁
For refraction at second surface (glass to air):
n/(-s₁' + t) + 1/s₂' = (1-n)/R₂
For thin lens, t → 0:
n/(-s₁') + 1/s₂' = (1-n)/R₂
Using object distance s = s₁ and image distance s' = s₂':
1/s + 1/s' = (n-1)[1/R₁ - 1/R₂]
Lens Equation: 1/f = 1/s + 1/s'
Comparing: 1/f = (n-1)[1/R₁ - 1/R₂]
This is the Lens Maker's Formula.
For convergent lens in air:
- f > 0: both surfaces converging
- R₁ > 0 (convex), R₂ < 0 (convex)
Q4. An object 5 cm tall is placed 20 cm from a convex lens of focal length 10 cm. Find image position, size, and nature.
Given:
Object height h₁ = 5 cm
Object distance u = 20 cm
Focal length f = 10 cm (convex lens)
Using lens equation: 1/f = 1/u + 1/v
1/10 = 1/20 + 1/v
1/v = 1/10 - 1/20 = 2/20 - 1/20 = 1/20
v = 20 cm
Image is at 20 cm from lens (on opposite side of object)
Magnification: m = -v/u = -20/20 = -1
Image height: h₂ = m × h₁ = (-1) × 5 = -5 cm
(Negative sign indicates inverted image)
Nature of image:
- Real image (v > 0)
- Inverted (m < 0)
- Same size as object (|m| = 1)
- At 2f position (u = 2f, v = 2f)
Answers:
- Image position: 20 cm from lens (opposite to object)
- Image height: 5 cm (inverted)
- Image nature: Real, inverted, same size
- Magnification: -1
Q5. What is power of a lens? A person has near point at 50 cm. What power of lens is required to correct this vision defect?
Power of Lens: Reciprocal of focal length in meters.
P = 1/f (in diopters, D)
f in meters
Vision Defect: Hypermetropia (far-sightedness)
Near point at 50 cm instead of normal 25 cm
Correction Required:
For comfortable reading at D = 25 cm (normal near point), the lens should form virtual image at 50 cm when object is at 25 cm.
Using lens equation: 1/f = 1/u + 1/v
Here: u = 25 cm = 0.25 m, v = -50 cm = -0.5 m (virtual image, negative)
1/f = 1/0.25 + 1/(-0.5)
1/f = 4 - 2 = 2
f = 0.5 m = +50 cm
Power: P = 1/f = 1/0.5 = +2 D
Answer:
- Power required: +2 D (diopters)
- Type of lens: Convex lens
- Formula: P = 1/f where f in meters
Note: Positive power indicates convergent (convex) lens needed for hypermetropia correction.
Q6. Explain the construction and working of a microscope. Derive the magnifying power.
Microscope: Optical instrument for viewing small objects magnified.
Construction:
- Objective lens: small focal length (f_o ≈ 1-5 mm), high power
- Eyepiece (ocular): moderate focal length (f_e ≈ 2-3 cm), acts as magnifying glass
- Tube length: distance between focal points of objective and eyepiece
- Lenses placed close together: separation ≈ L (tube length)
Working:
1. Object placed just beyond focal point of objective
2. Objective forms real, magnified, inverted image
3. This image acts as object for eyepiece
4. Eyepiece forms virtual, magnified image at infinity (for relaxed eye)
Magnifying Power:
For final image at infinity:
m = -(v_o/u_o) × (D/f_e)
Where: u_o = object distance from objective
v_o = image distance from objective
D = least distance of distinct vision (25 cm)
f_e = focal length of eyepiece
For object at focal point of objective (u_o ≈ f_o):
v_o ≈ L (tube length)
M = -(L/f_o) × (D/f_e) = -LD/(f_o f_e)
Magnifying power: M = |L/f_o| × |D/f_e|
For both lenses: M = M_o × M_e
Typical values: M = 10 × 40 = 400× to 1000×
Q7. Derive expression for resolving power of a microscope and explain its significance.
Resolving Power: Ability of microscope to distinguish between two closely spaced objects.
Rayleigh's Criterion: Two objects are just resolved when central maximum of one diffraction pattern coincides with first minimum of the other.
Minimum resolvable distance (Rayleigh limit):
δx = λ/(2 sin θ) or δx = λ/(2 NA)
Where NA = n sin θ = numerical aperture
n = refractive index of medium
θ = half-angle of cone of light from object
Resolving Power = 1/δx = 2 NA/λ = 2n sin θ/λ
For air: NA = sin θ ≈ 0.9 (maximum)
For oil immersion: n = 1.5, NA = 1.5 sin θ ≈ 1.35
Expression: δx = λ/(2n sin θ)
To maximize resolving power (minimize δx):
1. Decrease wavelength λ → use UV light
2. Increase refractive index n → use oil immersion
3. Increase NA → larger aperture
Significance:
- Determines magnification needed for useful viewing
- Useful magnification = 500 × to 1000 × resolving power
- Beyond this range, magnification becomes "empty magnification"
- For λ = 550 nm (visible): minimum resolvable distance ≈ 0.2 μm with oil immersion
Q8. A convex mirror has focal length 15 cm. An object is placed 30 cm from it. Find image position, size, and nature.
Given:
Convex mirror: f = -15 cm (focal length negative for convex mirror)
Object distance: u = -30 cm (object on front side, by convention)
Using sign convention: object distance negative
Actually, using proper sign convention:
u = 30 cm (object distance positive)
f = -15 cm (convex mirror, focal length negative)
Using mirror equation: 1/f = 1/u + 1/v
1/(-15) = 1/30 + 1/v
-1/15 = 1/30 + 1/v
1/v = -1/15 - 1/30 = -2/30 - 1/30 = -3/30 = -1/10
v = -10 cm
Image is 10 cm behind the mirror (virtual)
Magnification: m = -v/u = -(-10)/30 = 10/30 = 1/3
Image height: h' = m × h = (1/3) × h
(Erect image, 1/3 size of object)
Nature of image:
- Virtual (v < 0)
- Erect (m > 0)
- Diminished (|m| < 1)
- Always formed behind mirror for convex mirror
Answers:
- Image position: 10 cm behind mirror
- Magnification: +1/3
- Image nature: Virtual, erect, diminished
- Image height: 1/3 of object height
Ray Optics and Optical Instruments — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. An object is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. What is the image distance?
- A. -30 cm
- B. +30 cm
- C. -6 cm
- D. +6 cm
Answer: A. -30 cm — Using the mirror formula 1/f = 1/v + 1/u, with f = -10 cm and u = -15 cm, we get 1/v = 1/f - 1/u = 1/(-10) - 1/(-15) = -1/10 + 1/15 = (-3+2)/30 = -1/30. So, v = -30 cm.
MCQ 2. A ray of light travels from air into a medium with an angle of incidence of 45 degrees and an angle of refraction of 30 degrees. What is the refractive index of the medium?
- A. √2
- B. √3/2
- C. 1/√2
- D. 2/√3
Answer: A. √2 — According to Snell's Law, n1 sinθ1 = n2 sinθ2. Here, n1 (air) = 1, θ1 = 45°, θ2 = 30°. So, 1 * sin(45°) = n2 * sin(30°). This gives n2 = sin(45°)/sin(30°) = (1/√2) / (1/2) = 2/√2 = √2.
MCQ 3. Two thin convex lenses of focal lengths 20 cm and 40 cm are placed in contact. What is the power of the combination?
- A. +7.5 D
- B. +5 D
- C. +2.5 D
- D. +1.5 D
Answer: A. +7.5 D — For thin lenses in contact, the power of the combination P = P1 + P2. P1 = 1/f1 = 1/0.20 m = +5 D. P2 = 1/f2 = 1/0.40 m = +2.5 D. So, P = 5 + 2.5 = +7.5 D.
MCQ 4. For total internal reflection to occur, light must travel from:
- A. Denser to rarer medium
- B. Rarer to denser medium
- C. Air to glass
- D. Water to air
Answer: A. Denser to rarer medium — Total internal reflection occurs when light travels from a denser medium to a rarer medium, and the angle of incidence in the denser medium is greater than the critical angle.
MCQ 5. A plano-convex lens has a radius of curvature of 20 cm for its curved surface and a refractive index of 1.5. What is its focal length?
- A. 10 cm
- B. 20 cm
- C. 30 cm
- D. 40 cm
Answer: D. 40 cm — For a plano-convex lens, one surface is flat (R1 = infinity) and the other is curved (R2 = -20 cm for a convex surface refracting light). Using the lens maker's formula: 1/f = (n-1)(1/R1 - 1/R2). 1/f = (1.5-1)(1/∞ - 1/(-20)) = 0.5 * (0 + 1/20) = 0.5/20 = 1/40. So, f = 40 cm.
MCQ 6. Myopia is a defect of vision where a person cannot see:
- A. Nearby objects clearly
- B. Distant objects clearly
- C. Any object clearly
- D. Objects in dim light
Answer: B. Distant objects clearly — Myopia, or nearsightedness, is a condition where light entering the eye is focused in front of the retina, making distant objects appear blurry. It is corrected using a concave lens.
MCQ 7. An astronomical telescope has an objective lens of focal length 100 cm and an eyepiece of focal length 5 cm. What is its magnifying power in normal adjustment?
- A. 20
- B. 100
- C. 500
- D. 5
Answer: A. 20 — For an astronomical telescope in normal adjustment (image at infinity), the magnifying power M = -f_o / f_e. So, M = 100 cm / 5 cm = 20 (magnitude).
MCQ 8. A small air bubble is located at the center of a glass sphere of radius 4 cm and refractive index 1.5. Where will the image of the bubble appear when viewed from outside?
- A. At the center
- B. 2 cm from the surface
- C. 4 cm from the surface
- D. 6 cm from the surface
Answer: A. At the center — For refraction at a spherical surface, n2/v - n1/u = (n2 - n1)/R. Here, the object (bubble) is at the center, so u = -R. Light travels from glass (n1 = 1.5) to air (n2 = 1). The radius R is for the surface being viewed from, so R = -4 cm. 1/v - 1.5/(-4) = (1 - 1.5)/(-4). 1/v + 1.5/4 = -0.5/(-4) = 0.5/4. 1/v = 0.5/4 - 1.5/4 = -1/4. So v = -4 cm. The image appears at the center (same location as the object).
MCQ 9. Which of the following mirrors is used by dentists to see enlarged images of teeth?
- A. Plane mirror
- B. Convex mirror
- C. Concave mirror
- D. Parabolic mirror
Answer: C. Concave mirror — Concave mirrors produce enlarged and virtual images when the object is placed between the pole and the principal focus, making them suitable for dentists.
MCQ 10. When white light passes through a prism, which color deviates the most and which deviates the least?
- A. Red most, Violet least
- B. Violet most, Red least
- C. Green most, Yellow least
- D. Blue most, Orange least
Answer: B. Violet most, Red least — Due to dispersion, violet light has a shorter wavelength and higher refractive index in the prism material, causing it to deviate the most. Red light has a longer wavelength and lower refractive index, deviating the least.
MCQ 11. The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is:
- A. 0 degrees
- B. 45 degrees
- C. 90 degrees
- D. 180 degrees
Answer: C. 90 degrees — The critical angle is defined as the angle of incidence in the denser medium for which the refracted ray grazes the interface, meaning the angle of refraction in the rarer medium is 90 degrees.
MCQ 12. In a compound microscope, the focal lengths of the objective and eyepiece are 0.5 cm and 2 cm, respectively. If the object is placed 0.5675 cm from the objective, and the final image is formed at the least distance of distinct vision (25 cm), what is the magnifying power of the microscope?
- A. -150
- B. -120
- C. -100
- D. -75
Answer: C. -100 — For the objective lens: u_o = -0.5675 cm, f_o = 0.5 cm. 1/v_o = 1/f_o + 1/u_o = 1/0.5 + 1/(-0.5675) = 2 - 1.762 = 0.238. So v_o = 4.2 cm. Magnification m_o = v_o/u_o = 4.2/(-0.5675) ≈ -7.407. For the eyepiece, when the final image is at the least distance of distinct vision (D=25 cm), the magnifying power is m_e = (1 + D/f_e) = (1 + 25/2) = 13.5. Total magnifying power M = m_o * m_e = -7.407 * 13.5 ≈ -100.
MCQ 13. A ray of light passes through an equilateral glass prism with a refractive index of √3. What is the angle of minimum deviation?
- A. 30°
- B. 45°
- C. 60°
- D. 90°
Answer: C. 60° — For minimum deviation in an equilateral prism (A=60°), n = sin((A+Dm)/2) / sin(A/2). So, √3 = sin((60+Dm)/2) / sin(30°). √3 = sin((60+Dm)/2) / (1/2). sin((60+Dm)/2) = √3/2. So (60+Dm)/2 = 60°. 60+Dm = 120°. Dm = 60°.
MCQ 14. A person can read a book held at 25 cm, but cannot see objects beyond 1 meter. What power of lens is required to correct this defect and see distant objects clearly?
- A. +1 D
- B. -1 D
- C. +0.5 D
- D. -0.5 D
Answer: B. -1 D — The person has myopia (nearsightedness) with a far point of 1 meter (100 cm). To correct this, a concave lens is needed that forms a virtual image of an object at infinity at the person's far point. So, u = -∞, v = -100 cm. Using the lens formula, the focal length f = -100 cm = -1 m. The power P = 1/f = 1/(-1) = -1 D.
MCQ 15. In a lens experiment, an object and its real image are separated by a distance D. If the focal length of the lens is f, then the minimum distance D for which a real image is formed is:
- A. f
- B. 2f
- C. 3f
- D. 4f
Answer: D. 4f — For a real image to be formed by a convex lens, the object must be beyond the focal length. The distance between the object and its real image (D) is minimum when the object is placed at 2f, in which case the image is also formed at 2f on the other side. So u = 2f, v = 2f, and D = |u| + |v| = 2f + 2f = 4f.
MCQ 16. A point object is placed on the principal axis of a concave mirror of focal length 20 cm at a distance of 40 cm from it. If the object starts moving with a velocity of 2 cm/s towards the mirror, what is the initial velocity of the image?
- A. -2 cm/s
- B. +2 cm/s
- C. -4 cm/s
- D. +4 cm/s
Answer: A. -2 cm/s — Using the mirror formula, 1/f = 1/v + 1/u. Differentiating with respect to time: 0 = (-1/v^2) dv/dt + (-1/u^2) du/dt. So, dv/dt = -(v^2/u^2) du/dt = -m^2 du/dt. Given f = -20 cm, u = -40 cm. Since u = 2f, the object is at the center of curvature. For object at C, image is also at C, so v = -40 cm. Magnification m = -v/u = -(-40)/(-40) = -1. The object velocity du/dt = +2 cm/s (towards the mirror). So, dv/dt = -(-1)^2 * (+2) = -2 cm/s. The image also moves towards the mirror with the same speed.
MCQ 17. The refractive index of a medium is defined as the ratio of:
- A. Speed of light in vacuum to speed of light in medium
- B. Speed of light in medium to speed of light in vacuum
- C. Wavelength in vacuum to wavelength in medium
- D. Wavelength in medium to wavelength in vacuum
Answer: A. Speed of light in vacuum to speed of light in medium — The refractive index (n) of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in the medium (v), i.e., n = c/v.
MCQ 18. A coin is placed at the bottom of a beaker containing water (refractive index 4/3) up to a height of 12 cm. What is the apparent depth of the coin?
- A. 9 cm
- B. 12 cm
- C. 16 cm
- D. 18 cm
Answer: A. 9 cm — Apparent depth = Real depth / refractive index. Real depth = 12 cm, n = 4/3. Apparent depth = 12 / (4/3) = 12 * 3 / 4 = 9 cm.
MCQ 19. Convex mirrors are commonly used as rearview mirrors in vehicles because they:
- A. Produce real images
- B. Form magnified images
- C. Provide a wider field of view
- D. Are cheaper to manufacture
Answer: C. Provide a wider field of view — Convex mirrors always form virtual, erect, and diminished images, covering a wider field of view compared to plane mirrors, which is crucial for rearview applications.
MCQ 20. Two thin lenses of focal lengths 10 cm and -20 cm are placed in contact. What is the focal length of the combination?
- A. +10 cm
- B. -10 cm
- C. +20 cm
- D. -20 cm
Answer: C. +20 cm — For thin lenses in contact, 1/F = 1/f1 + 1/f2. So, 1/F = 1/10 + 1/(-20) = 1/10 - 1/20 = (2-1)/20 = 1/20. Therefore, F = +20 cm.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.