Region I and II are separated by a spherical surface of radius 25cm. An object is kept in region I at a distance of 40cm from the surface. The distance of the image from the surface is:

Option 1 -

9.52cm

Option 2 -

18 .23cm

Option 3 -

37.58cm

Option 4 -

55.44cm

6 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Since direction of incident ray (DIR) is from left to write, so considering refraction at point M, we can write

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

-> $\frac{1.4}{v} - \frac{1.25}{- 40} = \frac{1.4 - 1.25}{- 25}$

$\Rightarrow - \frac{1.4}{v} = \frac{0.15}{25} + \frac{1.25}{40} = \frac{1.2 + 6.25}{200} = \frac{7.45}{200}$

$\Rightarrow v = - \frac{200 \times 1.4}{7.45} = - 37.58 c m$

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Region I and II are separated by a spherical surface of radius 25cm. An object is kept in region I at a distance of 40cm from the surface. The distance of the image from the surface is:

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