One end of a cylindrical glass rod ends in a hemispherical surface of radius R = 20 mm. The rod is immersed in water of index 1.33. Find the image distance if object distance is 80 mm.
185

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Solution

The correct option is A 185
Equation of refraction of light at curved surfaces is given as:
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
$\frac{1.5}{v} - \frac{1.33}{- 80} = \frac{1.5 - 1.33}{20}$
$v \approx - 185 m m$

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One end of a cylindrical glass rod ends in a hemispherical surface of radius R = 20 mm. The rod is immersed in water of index 1.33. Find the image distance if object distance is 80 mm.185

The correct option is A 185Equation of refraction of light at curved surfaces is given as: μ2v−μ1u=μ2−μ1R 1.5v−1.33−80=1.5−1.3320 v≈−185mm

One end of a cylindrical glass rod ends in a hemispherical surface of radius R = 20 mm. The rod is immersed in water of index 1.33. Find the image distance if object distance is 80 mm.
185

The correct option is A 185
Equation of refraction of light at curved surfaces is given as:
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
$\frac{1.5}{v} - \frac{1.33}{- 80} = \frac{1.5 - 1.33}{20}$
$v \approx - 185 m m$