Sum

Locate the image formed by refraction in the situation shown in figure.

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#### Solution

Given,

Let the refractive indices of two mediums be μ_{1}=1.0 and μ_{2} =1.5

Point C is the centre of curvature, the distance between C and the pole is 20 cm.

Therefore, radius of curvature (*R)* = 20 cm

Distance between source S and pole is 25 cm.

Therefore, object distance (*u)* = −25

\[\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\]

\[\Rightarrow \frac{1 . 5}{v} - \frac{1}{( - 25)} = \frac{0 . 5}{20}\]

\[\Rightarrow \frac{1 . 5}{v} = \frac{- 3}{200}\]

\[\therefore v = - \frac{200 \times 1 . 5}{0 . 3 \times 10} = - 100\]

Hence, the required location of the image is 100 cm from the pole and on the side of S.

Concept: Refraction

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