
Given
u=−5cm
f=2−R=−240=−20cm
Now
v1+u1=f1
v1=−201=(−5)1
∴ v=+320cm
Then
Object distance d=5+320=335cm (below water surface)
The image after refraction
d′=d(μ1μ2)
=335(1.331)
=8.77
d′≃8.8cm
JEE Main 2019 — Physics Optics
A concave mirror has radius of curvature of 40cm. It is at the bottom of a glass that has water filled up to 5cm (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance d from the surface of water. The value of d is close to: (Refractive index of water =1.33)

Held on 12 Apr 2019 · Verified 6 Jul 2026.
13.4cm
8.8cm
6.7cm
11.7cm
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