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Power of a convex lens of refractive index 3/2 is 2.5 D in air. If it is inerted in liquid of refractive index 2, then new power will be ...... |
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Answer» -1.25 `THEREFORE 2.5=(3/2-1)((1)/(R_1)-(1)/(R_2))` … (1) for air Also,`(1)/(f.)=(""_lmu_g-1)((1)/(R_1)-(1)/(R_2))` [`""_lmu_g=(3/2)/(2)=3/4` `therefore (1)/(f.)=(3/4-1)((1)/(R_1)-(1)/(R_2))` …(2) By TAKING RATIO of equation (1) and (2), `2.5f.=(0.5)/(-0.25)` `therefore f.=(-5)/(25xx0.25)` `therefore(1)/(f.)=-1.25 D` |
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