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Question
Locate the image formed by refraction in the situation shown in figure.
Solution
Given,
Let the refractive indices of two mediums be μ1=1.0 and μ2 =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.
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