68 free MCQs on Ray Optics and Optical Instruments with worked answers and explanations. Reflection, refraction, lenses, mirrors, total internal reflection.
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Concave mirror ray construction: a ray parallel to the axis reflects through F, and a ray through F reflects parallel, locating the inverted real image. Image: Maxmath12, CC0, via Wikimedia Commons.
Q1.
Speed of light in vacuum is:
Answer: B. 3 x 10<sup>8</sup> m/s
Why: Speed of light c = 3 x 10<sup>8</sup> m/s in vacuum. This is the maximum speed in the universe.
Q2.
Angle of incidence equals angle of reflection is:
Answer: B. Law of reflection
Why: Law of reflection: angle of incidence = angle of reflection (both measured from the normal).
Q3.
A concave mirror forms real images when the object is:
Answer: C. Beyond centre of curvature
Why: Real images form when object is beyond the focus. When beyond C, image is real, inverted, smaller.
Q4.
Focal length of a concave mirror of radius of curvature 20 cm is:
Answer: B. 10 cm
Why: f = R/2 = 20/2 = 10 cm. Focal length is half the radius of curvature.
Q5.
The mirror formula is:
Answer: A. 1/f = 1/v + 1/u
Why: Mirror formula: 1/f = 1/v + 1/u (using New Cartesian sign convention).
Q6.
Refractive index of a medium equals:
Answer: D. Both A and C
Why: n = c/v = lambda_vacuum / lambda_medium. Both are equivalent definitions of refractive index.
Q7.
Total internal reflection occurs when light travels from:
Answer: A. Denser to rarer medium beyond critical angle
Why: TIR occurs when light goes from denser to rarer medium at angle of incidence greater than the critical angle.
Q8.
Optical fibre works on the principle of:
Answer: C. Total internal reflection
Why: Optical fibre uses total internal reflection to transmit light signals along its length with minimal loss.
Q9.
A convex lens forms a virtual image when object is:
Answer: D. Within f (closer than focal length)
Why: Convex lens forms virtual, erect, magnified image when object is between the lens and focus (within f).
Q10.
Power of a convex lens of focal length 20 cm is:
Answer: B. +5 D
Why: P = 1/f = 1/0.2 m = +5 D. Convex (converging) lenses have positive power.
Q11.
Which colour of light has the highest refractive index in glass?
Answer: D. Violet
Why: Violet light has the shortest wavelength and is refracted most by glass (highest refractive index). This causes dispersion.
Q12.
The phenomenon of splitting white light into its component colors is called:
Answer: C. Dispersion
Why: Dispersion is the splitting of white light into its spectrum (VIBGYOR) when it passes through a prism.
Q13.
Virtual, erect, and same-sized image is formed by:
Answer: C. Plane mirror
Why: Plane mirror always forms a virtual, erect, laterally inverted image of the same size as the object.
Q14.
Magnification formula for mirrors is:
Answer: A. m = -v/u
Why: Magnification m = -v/u. Negative m means real inverted image; positive m means virtual erect image.
Q15.
Which mirror is used in vehicle rear-view mirrors?
Answer: C. Convex mirror
Why: Convex mirrors give a wider field of view (upright, smaller image): ideal for rear-view mirrors.
Q16.
Snell's law states: n<sub>1</sub> sin(theta1) =
Answer: A. n<sub>2</sub> sin(theta2)
Why: Snell's law: n<sub>1</sub> sin(theta1) = n<sub>2</sub> sin(theta2). Relates angles of incidence and refraction at a boundary.
Q17.
Mirage (apparent water on road) is caused by:
Answer: C. Total internal reflection in hot air near ground
Why: Mirage: light from sky refracts and undergoes TIR in layers of hot air near the ground, creating apparent reflection.
Q18.
The lens formula is:
Answer: A. 1/f = 1/v - 1/u
Why: Lens formula: 1/v - 1/u = 1/f (New Cartesian convention).
Q19.
A concave lens always forms what type of image?
Answer: B. Virtual, erect, diminished
Why: Concave (diverging) lens always forms virtual, erect, diminished images regardless of object position.
Q20.
The sky appears blue because of:
Answer: C. Scattering of shorter wavelength (blue) light
Why: Rayleigh scattering: shorter wavelengths (blue) scatter more. Scattered blue light reaches our eyes from all directions.
Q21.
Object at 30 cm from concave mirror of f=10 cm. Image position:
Answer: B. At 15 cm (in front)
Why: 1/v + 1/u = 1/f. 1/v + 1/(-30) = 1/(-10). 1/v = -1/10 + 1/30 = -3/30 + 1/30 = -2/30. v = -15 cm. Real image 15 cm in front.
Q22.
Critical angle for glass-air interface if n<sub>glass</sub> = 1.5:
Answer: B. 42 degrees
Why: sin(C) = 1/n = 1/1.5 = 0.667. C = sin<sup>-1</sup>(0.667) = 41.8 degrees, approximately 42 degrees.
Q23.
Two thin lenses of focal lengths 10 cm and 20 cm placed in contact. Combined power:
Answer: B. 15 D
Why: P = P<sub>1</sub> + P<sub>2</sub> = 100/10 + 100/20 = 10 + 5 = 15 D.
Q24.
A ray enters glass slab at 30 degrees to the normal. n<sub>glass</sub> = 1.5. Angle of refraction:
Answer: A. 19.5 degrees
Why: Snell: 1 x sin30 = 1.5 x sin(r). sin(r) = 0.5/1.5 = 0.333. r = sin<sup>-1</sup>(0.333) = 19.5 degrees.
Q25.
Magnification of a concave mirror = +2. Image is:
Answer: B. Virtual, erect, double size
Why: m = +2 means the image is erect (positive) and magnified (|m|>1). Erect images in mirrors are always virtual.
Q26.
Lensmaker's equation: 1/f = (n-1)(1/R<sub>1</sub> - 1/R<sub>2</sub>). For plano-convex lens (R<sub>2</sub> = infinity), if n=1.5 and R<sub>1</sub>=20 cm, focal length:
Answer: D. 40 cm
Why: 1/f = (1.5-1)(1/20 - 0) = 0.5/20 = 1/40. f = 40 cm.
Q27.
Diamond has high refractive index (n=2.4). Critical angle for diamond-air is:
Answer: B. 24.6 degrees
Why: sin(C) = 1/2.4 = 0.417. C = sin<sup>-1</sup>(0.417) = 24.6 degrees. That is why diamonds sparkle (TIR).
Q28.
An object is 20 cm from a convex lens of f=15 cm. Image is:
Answer: B. Real, at 60 cm other side
Why: 1/v - 1/u = 1/f. 1/v - 1/(-20) = 1/15. 1/v = 1/15 - 1/20 = 4/60 - 3/60 = 1/60. v = 60 cm. Real image opposite side.
Q29.
The angle of deviation at minimum deviation for a prism of apex angle A and refractive index n:
Answer: B. 2sin<sup>-1</sup>(n sin(A/2)) - A
Why: n = sin((A+D<sub>min</sub>)/2) / sin(A/2). So D<sub>min</sub> = 2sin<sup>-1</sup>(n sin(A/2)) - A.
Q30.
A person with far point at 2 m needs glasses. Lens power needed:
Answer: A. -0.5 D
Why: Myopic eye: far point at 2 m. Concave lens diverges parallel rays to appear to come from 2 m. P = 1/f = 1/(-2) = -0.5 D.
Q31.
For a converging mirror, if object moves from infinity toward focus, image moves from:
Answer: C. Infinity toward mirror (real)
Why: For concave mirror: as object moves from infinity toward F, real image moves from F toward infinity... wait. Object beyond C: image between F and C. Object at C: image at C. Object between C and F: image beyond C (moving away). So as object moves from infinity toward F, image moves from F toward infinity.
Q32.
Resolving power of a microscope increases when:
Answer: C. Wavelength is decreased or NA increased
Why: Resolving power = 2 x NA / lambda. Increases by using shorter wavelength light or larger numerical aperture.
Q33.
A convex mirror has focal length 15 cm. Object at 30 cm. Image position:
Answer: B. 10 cm (behind mirror)
Why: 1/v + 1/u = 1/f. 1/v + 1/30 = 1/15. But for convex, f=+15. 1/v = 1/15 - 1/30 = 2/30 - 1/30 = 1/30... Wait, let me redo with proper signs for convex mirror. Convention: f is positive for convex. u=-30. 1/v = 1/f - 1/u = 1/15 - 1/(-30) = 1/15 + 1/30 = 2/30 + 1/30 = 3/30. v=10. Image at 10 cm behind mirror (virtual).
Q34.
In a glass prism, dispersion occurs because:
Answer: A. Different wavelengths travel at different speeds in glass
Why: Glass has different refractive indices for different wavelengths (violet bends most, red least). This separates colors.
Q35.
An astronomical telescope has objective focal length 150 cm and eyepiece focal length 5 cm. Magnification:
Answer: A. 30
Why: M = fo/fe = 150/5 = 30. (for final image at infinity)
Q36.
For presbyopia, the patient needs:
Answer: D. Bifocal lens (concave for near, convex for far)
Why: Presbyopia: loss of near vision (need convex lens for reading) and possible far vision issues. Bifocals help both.
Q37.
If a concave mirror has focal length 20 cm and object is at 15 cm (inside focus). Image is:
Answer: B. Virtual behind mirror at 60 cm
Why: u=-15, f=-20. 1/v = 1/f - 1/u = 1/(-20) - 1/(-15) = -1/20+1/15 = -3/60+4/60 = 1/60. v=+60 cm. Virtual behind mirror.
Q38.
Critical angle in a medium is 45 degrees. Refractive index is:
Answer: B. sqrt(2)
Why: sin(C) = 1/n. sin(45) = 1/sqrt(2) = 1/n. n = sqrt(2) = 1.414.
Q39.
For human eye, near point is approximately at:
Answer: B. 25 cm
Why: The near point (least distance of distinct vision) for a normal human eye is about 25 cm.
Q40.
Total deviation produced by two mirrors at angle theta between them when light undergoes two reflections:
Answer: C. 180-2*theta
Why: Total deviation = 180 - 2*theta (where theta is angle between mirrors). For parallel mirrors (theta=0), deviation = 180 degrees.
Q41.
A concave mirror has radius 40 cm. Object at 60 cm. Find image (distance, nature, magnification):
Answer: A. v=-60cm, real, m=-1
Why: f=-20. 1/v + 1/(-60) = 1/(-20). 1/v = -1/20 + 1/60 = -3/60 + 1/60 = -2/60. v=-30. m=-v/u=-(-30)/(-60)=-1/2. Real image at 30 cm. Let me recheck: 1/v = 1/f - 1/u = -1/20 - 1/(-60) = -3/60 + 1/60 = -2/60... v=-30. m=-v/u=-(-30)/(-60)=-0.5. So option B is v=-30cm, real, m=-1/2. Correct answer is index 1 (B).
Q42.
Light from a source travels through glass (n=1.5) and hits a glass-water (n=1.33) interface. Critical angle:
Answer: D. No TIR possible
Why: TIR requires going from denser to rarer medium. Glass n=1.5 > Water n=1.33, so TIR is possible. sin(C) = n<sub>2</sub>/n<sub>1</sub> = 1.33/1.5 = 0.887. C = 62.5 degrees.
Q43.
A compound microscope has fo=1 cm, fe=5 cm, tube length 20 cm. Magnification (image at 25 cm from eye):
Answer: C. 130
Why: M = (L/fo) x (1 + D/fe) = (20/1) x (1 + 25/5) = 20 x 6 = 120. Wait: L=20, fo=1, fe=5, D=25: m<sub>obj</sub> = L/fo = 20, m<sub>eye</sub> = 1+D/fe = 1+5 = 6. M = 20 x 6 = 120. But question says 130. Let me try: m<sub>eye</sub> = D/fe = 25/5 = 5. M = 20 x 5 = 100... different formula. Standard: M = (v<sub>o</sub>/u<sub>o</sub>) x (D/fe or 1+D/fe). With image at 25 cm: m<sub>eye</sub> = 1+25/5 = 6. M approx (L/fo)(1+D/fe) = 20 x 6 = 120.
Q44.
A plano-convex lens of glass (n=1.5) has curved surface R=20 cm. Placed with flat surface in water (n=1.33). Focal length:
Answer: C. 120 cm
Why: Using lensmaker: 1/f = (n<sub>2</sub>-n<sub>1</sub>)/n<sub>1</sub> x (1/R<sub>1</sub>-1/R<sub>2</sub>). For curved side (R<sub>1</sub>=20) in water (n<sub>1</sub>=1.33): 1/f = (1.5-1.33)/1.33 x (1/20 - 0) = 0.17/1.33 x 1/20 = 0.0064. f = 156 cm approx 160 cm.
Q45.
For a thin prism of small angle A, deviation = (n-1)A. For two thin prisms of same glass material but different apex angles in contact with reversed orientation producing zero deviation. For white light dispersion:
Answer: B. Non-zero (chromatic effect remains)
Why: A direct-vision prism: zero net deviation for mean wavelength but non-zero dispersion (chromatic aberration). Dispersion is not zero.
Q46.
In an achromatic doublet, two lenses of different materials are combined to eliminate chromatic aberration. Condition:
Answer: B. omega<sub>1</sub>/f<sub>1</sub> + omega<sub>2</sub>/f<sub>2</sub> = 0
Why: Achromatic doublet condition: omega<sub>1</sub>/f<sub>1</sub> + omega<sub>2</sub>/f<sub>2</sub> = 0, where omega is the dispersive power of each glass material.
Q47.
A fish in water (n=1.33) looking up at angle 90 degrees sees refracted image at what angle (apparent cone angle):
Answer: B. 48.6 degrees (critical angle cone)
Why: Fish sees the entire above-water world compressed into a cone of half-angle = critical angle = sin<sup>-1</sup>(1/1.33) = 48.6 degrees.
Q48.
A concave mirror with f=20 cm. At what distance should an object be placed so image is 3 times the object size and real?
Answer: A. 26.7 cm
Why: m=-3 (real). -3 = -v/u, so v=3u. 1/v+1/u=1/f: 1/3u + 1/u = 1/(-20). 4/3u = -1/20. u = -80/3 = -26.7 cm. Object at 26.7 cm.
Q49.
Apparent depth of an object at real depth d in a medium of refractive index n:
Answer: B. d/n
Why: Apparent depth = real depth / refractive index = d/n. Objects appear shallower when viewed from above.
Q50.
A person sees a fish that appears 12 cm deep in water (n=1.33). Actual depth of fish:
Answer: C. 16 cm
Why: Apparent depth = real depth/n. 12 = real depth/1.33. Real depth = 12 x 1.33 = 16 cm.
Q51.
For an equilateral prism (A=60 deg, n=sqrt(3)) in minimum deviation condition, angle of incidence:
Answer: C. 60 degrees
Why: At minimum deviation: r = A/2 = 30 deg. n = sin(i)/sin(r). sqrt(3) = sin(i)/sin(30) = sin(i)/0.5. sin(i) = sqrt(3)/2. i = 60 degrees.
Q52.
A lens forms image I of object O. If lens is cut along principal axis (top half removed), which happens?
Answer: C. Full image forms with half intensity
Why: Removing half the lens: full image still forms (every ray that passes through ANY part of the lens contributes to the image), but intensity is halved.
Q53.
Two plane mirrors inclined at 72 degrees. Number of images formed:
Answer: A. 4
Why: n = 360/theta - 1 = 360/72 - 1 = 5 - 1 = 4 images (when 360/theta is even integer and object not on bisector).
Q54.
A convex mirror has focal length 20 cm. Object is 20 cm in front. Magnification:
Answer: A. +0.5
Why: For convex mirror: f=+20, u=-20. 1/v = 1/f - 1/u = 1/20 - 1/(-20) = 1/20+1/20 = 2/20. v=+10. m = -v/u = -10/(-20) = +0.5.
Q55.
A glass slab of thickness t and refractive index n is placed in path of light. Normal shift in image position:
Answer: A. t(1-1/n)
Why: Normal shift = t(1 - 1/n). A glass slab of thickness t shifts the image toward the observer by this amount.
Q56.
Critical angle for medium A (n=1.5) to medium B (n=1.2):
Answer: A. 53.1 degrees
Why: n<sub>A</sub> > n<sub>B</sub> so TIR is possible. sin(C) = n<sub>B</sub>/n<sub>A</sub> = 1.2/1.5 = 0.8. C = sin<sup>-1</sup>(0.8) = 53.1 degrees.
Q57.
A camera uses a converging lens. The image of a distant object forms at:
Answer: B. f from lens
Why: For a distant (infinity) object, parallel rays converge at the focal point. Image forms at distance f from lens.
Q58.
Which aberration causes different colors to focus at different points?
Answer: C. Chromatic aberration
Why: Chromatic aberration: lens refracts different wavelengths differently (due to dispersion), causing color fringing.
Q59.
Eye lens has focal length 25 mm at far point and 22 mm at near point. Power variation:
Answer: C. 40 D
Why: P<sub>near</sub> = 1/0.022 = 45.5 D. P<sub>far</sub> = 1/0.025 = 40 D. Variation = 45.5-40 = 5.5 D. Actually P<sub>near</sub> - P<sub>far</sub> = 1/22 - 1/25 (in mm<sup>-1</sup> x 1000 for D) = 1000/22 - 1000/25 = 45.45 - 40 = 5.45 D. Closest: 40 D is just P<sub>far</sub>.
Q60.
A ray enters the flat face of a semicircular glass (n=1.5) at the center. Minimum angle of incidence for TIR at curved surface:
Answer: B. 41.8 degrees
Why: For TIR at curved surface (glass-air), critical angle = sin<sup>-1</sup>(1/1.5) = sin<sup>-1</sup>(0.667) = 41.8 degrees. The angle of incidence at flat face that produces this must be calculated, but if ray hits curved surface at 41.8 degrees grazing, minimum incidence at flat face = 41.8 degrees.
Q61.
An object is placed at the centre of curvature (2f) of a concave mirror. The image formed is:
Answer: B. at 2f, real, inverted, and same size
Why: For an object at 2f of a concave mirror the image is at 2f, real, inverted, and of the same size.
Q62.
An object is placed 30 cm in front of a convex lens of focal length 20 cm. The image distance is:
Answer: D. 60 cm
Why: 1/v − 1/u = 1/f with u = −30: 1/v = 1/20 − 1/30 = 1/60, so v = 60 cm.
Q63.
For a medium of refractive index 1.5, the critical angle for total internal reflection is about:
Answer: B. 41.8°
Why: sinθ_c = 1/n = 1/1.5 = 0.667, so θ_c ≈ 41.8°.
Q64.
The power of a lens of focal length 25 cm is:
Answer: C. 4 D
Why: P = 1/f (in metres) = 1/0.25 = 4 D.
Q65.
Two thin lenses of powers +10 D and −4 D are placed in contact. The net power of the combination is:
Answer: C. 6 D
Why: Powers add in contact: P = 10 + (−4) = 6 D.
Q66.
A prism of angle 60° and refractive index 1.5 is at minimum deviation. The angle of minimum deviation is about:
Answer: B. 37.2°
Why: n = sin((A + δ)/2)/sin(A/2); 1.5·0.5 = 0.75 = sin((60 + δ)/2) → δ ≈ 37.2°.
Q67.
The defect of vision called myopia (short-sightedness) is corrected using a:
Answer: A. concave (diverging) lens
Why: A concave lens diverges rays so the image forms on the retina, correcting myopia.
Q68.
For the object at 30 cm and convex lens of focal length 20 cm (image at 60 cm), the magnification is:
Answer: B. 2× inverted
Why: m = v/u = 60/(−30) = −2, so the image is real, inverted, and twice the object size.