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InterviewSolution
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The radiiof curvature of two surfacesof a convex lensis 0.2 m and 0.22 m. Find thefocal length of the lensif refractiveindexof the materialof lens is 1.5 . Also find thechange in focal length, if it is immersedin water of refractiveindex1.33. |
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Answer» SOLUTION :Given :`R_1+0.2m`, `R_2=-0.22`m , `n=1.5 , n_W=1.33` W.k.t.`1/t=(n_2-1)(1/R_1-1/R_2)` i.e., `1/f =(1.5-1)(1/0.2+1/0.22)` i.e.,`1/f=0.5xx(5+4.55)`=4.78 Hence `f=1/4.78`=0.209 m but `f_(w)/(f_a)=(n_r-1)/(._ln_g-1)` i.e, `f_(w)/(f_a)=(1.5-1)/(1.5/1.33-1)=3.91` i.e., `f_w`=3.91`f_n` `=3.91xx0.209` =0.817 m Focal lengthof lens in water = 0.817 m |
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