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A thin glass (refractive index 1.5) lens has optical power of `-5D` in air. Its optical power in a liquid medium with refractive index 1.6 will beA. 1DB. `-1D`C. `25 D`D. `-25D` |
Answer» Correct Answer - A Power of lens = `mu_(2) - mu_(1) ((1)/(R_(t) - (1)/(R_(2)))` in air power is -5D `therefore -5 = (1.5 -1) ((1)/(R_(1)) - (1)/(R_(2)))` ……. (i) In other medium power is P `P = (1.5 - 1.6) ((1)/(R_(1))- (1)/(R_(2)))` ……. (ii) Dividing (ii) by (i) `(P)/(-5) = (-0.1)/(1) xx (1)/(0.5) implies P = 1D` |
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