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A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a

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Convergent lens of focal length 3.5 R

Divergent lens of focal length 3.0 R

Convergent lens of focal length 3.0 R

Divergent lens of focal length 3.5 R

answer is A.

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Detailed Solution

$\frac{1}{f} = (μ_{g} - 1) \frac{2}{R} = (1.5 - 1) \frac{2}{- R} = - \frac{1}{R} f = - R - - - (1)$

$\frac{f^{1}}{f} = \frac{(μ_{g} - 1) μ_{L}}{μ_{g} - μ_{L}} \frac{f^{1}}{R} = \frac{(1.5 - 1) 1.75}{1.5 - 1.75} f' = + 3.5 R + c o n v e r g i n g l e n s$

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