JEE MainPhysicsModerateRay Optics

A concave mirror of radius of curvature 40 cm is filled with
water (n=4/3) upto a height of 12 cm. A point object O is kept on the
principal axis of the mirror at height 13.5 cm from the water surface.
The final image formed after refraction at water surface, reflection at
mirror and again refraction at water surface in succession is situated
at.

Show answer and solution

Refraction -1: Apparent depth from water surface

d^{'} = 13.5 \times \frac{4}{3} = 18cm

Reflection from concave mirror.

\frac{1}{v} + \frac{1}{( - 30)} = \frac{1}{- 20}

v = - 60cm

Image is formed 48 cm above water surface
Refraction-2:

d^{''} = \frac{48}{4/3} = 36cm

above water surface

Source: Grass-G / P-17 Physics paper 26 Nov Second.docx

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