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JEE • PHYSICS • RAY OPTICS

Ray optics practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard

1 of 8

Application +4 | -1

The exposure time of a camera lens at the

\frac{f}{{2.8}}

setting is

\frac{1}{{200}}

second. The correct time of exposure at

\frac{f}{{5.6}}

View answer and explanation

Correct option: B

0.02 sec

Option B (0.02 sec) is correct because the exposure time is proportional to the square of the f-number.

2 Hard

2 of 8

Application +4 | -1

A ray of light enters a rectangular glass slab of refractive index

\sqrt{3}

at angle of incidence

60^{\circ}

. It ravels a distance of

10 \mathrm{~cm}

inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is

View answer and explanation

Correct option: D

5 \mathrm{~cm}

Option D (

5 \text{ cm}

) is correct because it's calculated using the formula for lateral shift, considering the angle of incidence, angle of refraction, and thickness of the slab.

3 Hard

3 of 8

Application +4 | -1

For an object placed at a distance

2.4 \mathrm{~m}

from a lens, a sharp focused image is observed on a screen placed at a distance

12 \mathrm{~cm}

form the lens. A glass plate of refractive index

1.5

and thickness 1

\mathrm{cm}

is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

View answer and explanation

Correct option: B

3.2 m

Option B (3.2 m) is correct because it accounts for the shift in image position due to the glass plate and recalculates the object distance to maintain a focused image.

4 Hard

4 of 8

Application +4 | -1

The exposure time of a camera lens at the

\frac{f}{{2.8}}

setting is

\frac{1}{{200}}

second. The correct time of exposure at

\frac{f}{{5.6}}

View answer and explanation

Correct option: B

0.02 sec

Option B (0.02 sec) is correct because the exposure time is proportional to the square of the f-number.

5 Hard

5 of 8

Application +4 | -1

A beam of light of width

t

is incident on an air-water boundary at an angle of incidence

45^{\circ}

. The width of the beam in water is (Refractive index of water

=\mu

View answer and explanation

Correct option: C

\frac{\sqrt{2 \mu^2-1}}{\mu} \cdot t

Option C (

\frac{\sqrt{2 \mu^2-1}}{\mu} \cdot t

) is correct because it is derived from Snell's law and trigonometric relationships to find the width of the beam in water.

6 Hard

6 of 8

Application +4 | -1

For an object placed at a distance

2.4 \mathrm{~m}

from a lens, a sharp focused image is observed on a screen placed at a distance

12 \mathrm{~cm}

form the lens. A glass plate of refractive index

1.5

and thickness 1

\mathrm{cm}

View answer and explanation

Correct option: B

3.2 m

Option B (3.2 m) is correct because it accounts for the shift in image position due to the glass plate and recalculates the object distance to maintain a focused image.

7 Hard

7 of 8

Application +4 | -1

A concave mirror has a focal length

20 \mathrm{~cm}

. The distance between the two positions of the object for which the image size is double of the object size is

View answer and explanation

Correct option: A

20 \mathrm{~cm}

The correct answer is A (20 cm) because the distance between the two object positions is the difference between the object distances for real and virtual images with a magnification of 2.

8 Hard

8 of 8

Application +4 | -1

A ray enters a glass slab at an angle

\alpha

from air and the refractive index varies along the slab as

\mu=\mu_0-k t^2

where

t

is the distance measured along the normal.

\mu_0

and

k

are positive constants. If the glass slab is sufficiently thick, how far along the normal will the ray go before it reflects back?

View answer and explanation

Correct option: C

\sqrt{\frac{\mu_0-\sin \theta}{k}}

Option C (

\sqrt{\frac{\mu_0-\sin \theta}{k}}

) is correct because the distance is found by setting the refractive index at the reflection point equal to the sine of the incident angle.

Revision notes and key takeaways

Review RAY OPTICS in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 8 single-correct-answer MCQs on RAY OPTICS.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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Ray optics practice · Hard

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