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Question

In the figure, light is incident on the thin lens as shown. The radius of curvature for both the surface is R. The focal length of this system is

$f = \frac{μ_{3} R}{μ_{3} - μ_{1}}$
$f = \frac{μ_{2} R}{μ_{1} - μ_{2}}$
$f = \frac{μ_{3}}{μ_{2} - μ_{1}}$
$f = \frac{μ_{1} R}{μ_{1} - μ_{2}}$

f = \frac{μ_{2} R}{μ_{1} - μ_{2}}

f = \frac{μ_{3}}{μ_{2} - μ_{1}}

f = \frac{μ_{3} R}{μ_{3} - μ_{1}}

f = \frac{μ_{1} R}{μ_{1} - μ_{2}}

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Solution

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Consider refraction from the two spherical surfaces.
At surface 1,
$\frac{μ_{2}}{v_{1}} - \frac{μ_{1}}{\infty} = \frac{μ_{2} - μ_{1}}{R}$

At surface 2,
$\frac{μ_{3}}{v_{2}} - \frac{μ_{2}}{v_{1}} = \frac{μ_{3} - μ_{2}}{R}$

$⟹ v_{2} = \frac{μ_{3} R}{μ_{3} - μ_{1}} = f$

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