Question:
Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :
Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :
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Use \(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\) with \(\mu_1 = 1.25\), \(\mu_2 = 1.4\); the centre of curvature C lies on the incident side, so \(R\) is negative.
Updated On: Nov 4, 2026
- 55.44 cm
- 18.23 cm
- 9.52 cm
- 37.58 cm
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The Correct Option is D
Solution and Explanation
Light travels from region I (\(\mu_1 = 1.25\)) to region II (\(\mu_2 = 1.4\)). Take the direction of light as positive: \(u = -40\) cm, and since the centre C lies on the side of the object, \(R = -25\) cm.
\(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\Rightarrow \frac{1.4}{v} + \frac{1.25}{40} = \frac{0.15}{-25}\).
\(\frac{1.4}{v} = -0.006 - 0.03125 = -0.03725\), so \(v = -\frac{1.4}{0.03725} \approx -37.58\) cm.
The image is virtual, formed on the side of region I, at a distance of 37.58 cm from the surface.
Answer: (4) 37.58 cm
\(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\Rightarrow \frac{1.4}{v} + \frac{1.25}{40} = \frac{0.15}{-25}\).
\(\frac{1.4}{v} = -0.006 - 0.03125 = -0.03725\), so \(v = -\frac{1.4}{0.03725} \approx -37.58\) cm.
The image is virtual, formed on the side of region I, at a distance of 37.58 cm from the surface.
Answer: (4) 37.58 cm
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