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The focal length of a thin lens in vacuum is `f`. If the material of the lens has `mu = 3//2`, its focal length when immersed in water of refractive index `4//3` will be.A. `f`B. `4 f//3`C. `2 f`D. `4 f` |
Answer» Correct Answer - D `(1)/(f_(a)) = ((mu_(g))/(mu_(a)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` …(i) `(1)/(f_(w)) = ((mu_(g))/(mu_(w)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` …(ii) Dividing (i) by (ii), we get `(f_(w))/(f_(a)) = ((mu_(g)//mu_(a) - 1))/((mu_(g)//mu_(w) - 1)) = ((3//2 - 1))/(((3//2)/(4//3) -1)) = (1//2)/(1//8) = 4` `:. f_(w) = 4 f_(a) = 4 f` |
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