A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is (C is the center of curvature of the spherical surface and R is the radius of curvature) [2025]

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Q.

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is (C is the center of curvature of the spherical surface and R is the radius of curvature) [2025]

1 0.24 m right to the spherical surface

2 0.4 m left to the spherical surface

3 0.24 m left to the spherical surface

4 0.4 m right to the spherical surface

Ans.
(2)

For spherical surfaces

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

$\frac{1.5}{v} - \frac{1}{(- 0.2)} = \frac{1.5 - 1}{0.4}$

$\frac{1.5}{v} = \frac{0.5}{0.4} - \frac{1}{0.2}$

$\Rightarrow v = - 0.4 m$

Solution

Q.

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is (C is the center of curvature of the spherical surface and R is the radius of curvature) [2025]

1 0.24 m right to the spherical surface

2 0.4 m left to the spherical surface

3 0.24 m left to the spherical surface

4 0.4 m right to the spherical surface

Ans.
(2)

For spherical surfaces

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

$\frac{1.5}{v} - \frac{1}{(- 0.2)} = \frac{1.5 - 1}{0.4}$

$\frac{1.5}{v} = \frac{0.5}{0.4} - \frac{1}{0.2}$

$\Rightarrow v = - 0.4 m$

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Solution

Q. A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is (C is the center of curvature of the spherical surface and R is the radius of curvature) [2025]

1 0.24 m right to the spherical surface

2 0.4 m left to the spherical surface

3 0.24 m left to the spherical surface

4 0.4 m right to the spherical surface

Ans. (2) For spherical surfaces μ2v–μ1u=μ2–μ1R 1.5v–1(–0.2)=1.5–10.4 1.5v=0.50.4–10.2 ⇒ v=–0.4 m

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Q.

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is (C is the center of curvature of the spherical surface and R is the radius of curvature) [2025]

Ans.
(2)

For spherical surfaces

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

$\frac{1.5}{v} - \frac{1}{(- 0.2)} = \frac{1.5 - 1}{0.4}$

$\frac{1.5}{v} = \frac{0.5}{0.4} - \frac{1}{0.2}$

$\Rightarrow v = - 0.4 m$