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A double convex lens of radius of curvature of each surface equal to 20 cm is made of glass of refractive index 1.5. Find the ratio of the powers of the lens when placed in air to its power when placed in water. Take refractive index of water = 1.33. |
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Answer» Solution :`"Given, first radius of CURVATURE, "R_(1)=20cm` `"Second radius of curvature, "R_(2)=-20cm` `"Refractive index of lens, "mu_(g)=1.5` `"Refractive index of water, "mu_(e)=1.33` For the lens placed in AIR : `P_(1)=(mu_(g)-1)[(1)/(R_(1))-(1)/(R_(2))]` `=(1.5-1)[(1)/(0.2)-((-1)/(0.20))]` `=0.5xx(2)/(0.2)=5D` For the lens placed in LIQUID : `P_(2)=((u_(g))/(mu_(e))-1)[(1)/(R_(1))-(1)/(R_(2))]` `=((1.5)/(1.33)-1)[(1)/(0.2)-((-1)/(0.20))]` `=0.128xx(2)/(0.2)=1.28D` `therefore""(P_(1))/(P_(2))=(5)/(1.28)=3.9:1` |
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