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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSMedium

Question

An object is placed at a distance of 40 cm from a concave mirror of a focal length of 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be:

A

30 cm away from the mirror

B

36 cm away from the mirror

C

30 cm towards the mirror

D

36 cm towards the mirror

Step-by-Step Solution

We use the mirror formula: 1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f}.

Case 1 (Initial Position): Object distance, u1=−40 cmu_1 = -40 \text{ cm}. Focal length, f=−15 cmf = -15 \text{ cm} (concave mirror). 1v1=1f−1u1=1−15−1−40=−8+3120=−5120\frac{1}{v_1} = \frac{1}{f} - \frac{1}{u_1} = \frac{1}{-15} - \frac{1}{-40} = \frac{-8 + 3}{120} = \frac{-5}{120}. v1=−24 cmv_1 = -24 \text{ cm}. The initial image is 24 cm in front of the mirror.

Case 2 (Final Position):

  • The object is displaced 20 cm towards the mirror. New object distance u2=−(40−20)=−20 cmu_2 = -(40 - 20) = -20 \text{ cm}. 1v2=1f−1u2=1−15−1−20=−4+360=−160\frac{1}{v_2} = \frac{1}{f} - \frac{1}{u_2} = \frac{1}{-15} - \frac{1}{-20} = \frac{-4 + 3}{60} = \frac{-1}{60}. v2=−60 cmv_2 = -60 \text{ cm}. The new image is 60 cm in front of the mirror.

Displacement: Displacement = ∣v2∣−∣v1∣=60−24=36 cm|v_2| - |v_1| = 60 - 24 = 36 \text{ cm}. Since the image distance increased from 24 cm to 60 cm, the image moved away from the mirror.

Thus, the displacement is 36 cm away from the mirror.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from RAY OPTICS AND OPTICAL INSTRUMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSobjectplaceddistanceconcavemirror

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