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An equibiconvex lens of radius of curvature 0.20 and refractive index 1.5 immersed half inside water of RI 4//3 and the rest outside in air. A parallel beam of light in air is incident on it. Find the final position of the image. |
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Answer» Solution :GIVEN`r=0.20m ` `n_(w)=(4)/(3)=1.333` `n_(G)=1.50` `u=oo`  For reflection through I surface `n_(0)=1, n_(1)=1.50, u=oo, v=?` By using `(n_(0))/(-u)+(n_(1))/(v)=(n_(0)-n_(1))/(r)` i.e., `(1)/(oo)+(1.5)/(v)=(1.5-1)/(0.2)` `(1.5)/(v)=(0.5)/(0.2)=2.5` `thereforev=(1.5)/(2.5)=0.6m` Real image serves as virtual object for II surface. Put `u.=0.6m` `(n_(g))/(-0.6)+(n_(w))/(v)=(n_(g)-n_(w))/(r)` `therefore (1.5)/(-0.6)+(1.333)/(v)=(1.5-1.333)/(0.2)` `-2.5+(1.333)/(v)=(0.1667)/(0.2)=0.8335` `(1.333)/(v)=3.333` `v=(1.333)/(3.333)=0.3999` `v=0.40m` from the SECOND surface in the direction of INCIDENT light. |
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