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A thin transparent box is separated in two parts of $μ_{1} and μ_{2}$ by a spherical surface of radius of curvature R, as shown.

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For, $μ_{1} < μ_{2}$ , it behaves as a convergent lens

For, $μ_{1} > μ_{2}$ ,it behaves as a convergent lens

For, $μ_{1} < μ_{2}$ it behaves as a divergent lens

For, $μ_{1} = μ_{2}$ it behaves as a slab

answer is A, B, C.

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Detailed Solution

Equivalent focal length , f = $\frac{R}{μ_{1} - μ_{2}} .$

For converging lens f $> 0, f o r d i v e r g i n g l e n s f < 0 a n d f o r s l a b f = 0 .$

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