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A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water filled up to 5 cm (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)

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11.7 cm

13.4 cm

8.8 cm

6.7 cm

answer is C.

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Detailed Solution

If v is the distance of image formed by mirror

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ (or) $\frac{1}{v} + \frac{1}{- 5} = \frac{1}{- 20} ∴ v = \frac{20}{3} cm$

So the image formed below the mirror will act as the object for refraction at the water surface

$Image distance = \frac{20}{3} + 5 = \frac{35}{3} cm$

$∴ Apperent depth = \frac{Image distance}{μ} = \frac{(35 / 3)}{1 .33} = 8.75 cm ≃ 8 .8 cm$

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