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A convex lens of focal length 0.5m in air has a refractive index of 1.5.What will be its focal length when immersed in water of refractive index 1.33 ? |
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Answer» Solution :`(1)/(f) = ( N -1) ((1)/( R_(1)) +( 1)/( R_(2)))` For air `(1)/(0.5) = ( 1.5 -1) ((1)/( R_(1)) + (1)/( R_(2)))`....(i) For WATER `(1)/( f_(W)) = ((n_(g)) /(n_(w)) - 1)((1)/( R_(1)) + (1)/(R_(2)))` `(1)/(f_(w)) = ((1.5)/(1.33)-1) ((1)/(R_(1)) +(1)/(R_(2)))`....(2) DIVIDING equation(1) by (2) `f_(w) = 1.956m` |
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