The image for the converging beam after refraction through the curved surface (in the given figure) is formed at: (consider O to be origin)

Question

TestHub · Accepted Answer

The converging beam would meet at P, 30 cm to the right of O, so the surface sees a virtual object: $u = + 30 cm$

Refraction from glass (μ = 3/2) to air (μ = 1), R = +20 cm:

$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R} \Rightarrow \frac{1}{v} - \frac{3/2}{30} = \frac{1 - 3/2}{20} = - \frac{1}{40}$

$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \Rightarrow v = + 40 cm$

Real image at x = 40 cm. Hence option (A).

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