The image for the converging beam after refraction through the curved surface is formed at

40 c m

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\frac{40}{3} c m

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20 c m

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\frac{180}{7} c m

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Solution

The correct option is A $40 c m$
It is clear that the object is virtual and the image is real.

$\frac{n_{i}}{v} - \frac{n_{o}}{u} = \frac{n_{i} - n_{0}}{R} \frac{1}{v} - \frac{\frac{3}{2}}{(+ 30)} = \frac{1 - \frac{3}{2}}{+ 20} \frac{1}{v} = \frac{1}{20} - \frac{1}{40} \Rightarrow v = 40 c m$

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In the given figure, the radius of curvature of curved surface for both the planoconvex and plano-concave lens is 10 cm and refractive index for both is 1.5. The location of the final image from the plano-concave lens after all the refractions through lenses is

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The image for the converging beam after refraction through the curved surface is formed at

The correct option is A 40 cmIt is clear that the object is virtual and the image is real. niv−nou=ni−n0R1v−32(+30)=1−32+201v=120−140⇒v=40 cm

The correct option is A $40 c m$
It is clear that the object is virtual and the image is real.

$\frac{n_{i}}{v} - \frac{n_{o}}{u} = \frac{n_{i} - n_{0}}{R} \frac{1}{v} - \frac{\frac{3}{2}}{(+ 30)} = \frac{1 - \frac{3}{2}}{+ 20} \frac{1}{v} = \frac{1}{20} - \frac{1}{40} \Rightarrow v = 40 c m$