BITSAT PYQs for Reflection of Light by Spherical Mirrors with Solutions: Practice BITSAT Previous Year Questions

Updated 3+ months ago

Reflection of Light by Spherical Mirrors is an important topic in the Physics section in BITSAT exam. Practising this topic will increase your score overall and make your conceptual grip on BITSAT exam stronger.

This article gives you a full set of BITSAT PYQs for Reflection of Light by Spherical Mirrors with explanations for effective preparation. Practice of BITSAT Physics PYQs including Reflection of Light by Spherical Mirrors questions regularly will improve accuracy, speed, and confidence in the BITSAT 2026 exam.

BITSAT PYQs for Reflection of Light by Spherical Mirrors with Solutions

1.
In a Young's double slit experiment, the intensities at two points, for the path difference $ \frac{\lambda}{4} $ and $ \frac{\lambda}{3} $ ($ \lambda $ being the wavelength of light used) are $ I_1 $ and $ I_2 $ respectively. If $ I_0 $ denotes the intensity produced by each one of the individual slits, then $ \frac{I_1 + I_2}{I_0} $ is equal to:
- $ 3 $
- $ 5 $
- $ 7 $
- $ 10 $
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2.
A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of $9 \,kg$ is suspended from the wire. When this mass is replaced by a mass $M$, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of $M$
- 25 kg
- 5 kg
- 12.5 kg
- 1/25 kg
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3.
An object is placed at a distance $20\, cm$ from the pole of a convex mirror of focal length $20\, cm$. The image is produced at :
- 13.3 cm
- 20 cm
- 25 cm
- 10 cm
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4.
A simple pendulum doing small oscillations at a place $ R $ height above the Earth's surface has a time period of $ T_1 = 4 \, {s} $. $ T_2 $ would be its time period if it is brought to a point which is at a height $ 2R $ from the Earth's surface. Choose the correct relation [$ R $ = radius of Earth]:
- $ T_1 = T_2 $
- $ 2T_1 = 3T_2 $
- $ 3T_1 = 2T_2 $
- $ 2T_1 = T_2 $
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5.
A spherical ball of mass $ 20 $ kg is stationary at the top of a hill of height $ 100 $ m. It rolls down a smooth surface to the ground, then climbs up another hill of height $ 30 $ m and finally rolls down to a horizontal base at a height of $ 20 $ m above the ground. The velocity attained by the ball is:
- $ 20 $ m/s
- $ 40 $ m/s
- $ 10\sqrt{30} $ m/s
- $ 10 $ m/s
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