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Karnataka Board PUCPUC Science Class 11

P Locate the Image Formed by Refraction in the Situation Shown in Figure.

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Question

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

Sum
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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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Chapter 18: Geometrical Optics - Exercise [Page 414]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 18 Geometrical Optics
Exercise | Q 37 | Page 414
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