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A convex lens of focal length 0.24 m and of refractive index 1.5 is completely immersed in water of refractive 1.33. Find the changes in the focal length of the lens. |
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Answer» Solution :Lens in air, `(1)/(f_(G)) \= ( n _(g)-1) ((1)/(R_(1))-(1)/(R_(2)) )`, `(f_(g) = `FOCAL length in air ) `(1)/(0.24) = ( 1.5-1) ((1)/(R_(1))-(1)/(R_(2)))` Lens in water, ` (1)/( f_(gw)) = ( (n_(g))/(n_(w))-1)= ( (1)/( R_(1)) - ( 1)/( R_(2)))`,`( f_(gw)` = focal length of lens in water ) `(1)/( f_(gw)) = ((1.5)/(1.3) -1) ( (1 )/( R_(1))- ( 1)/( R_(2))) `....(ii) DIVIDING EQUATION (1) by (2), `f_(gw) = 0.939m` `:.` Change in focal length `= 0.699m ( with unit ) |
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