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A parallel beam of light strikes the first surface of a glass sphere of R.I 1.5 and radius of curvature 0.10 m. Find the positionof the final image. |
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Answer» <P> SOLUTION :`u=OO``n_(g)=1.5` `n_(a)=1` `r=0.10m` (i) for I surface using the FORMULA `(n_(0))/(-u)+(n_(1))/(v)=(n_(0)-n_(1))/(r)` `(1)/(oo) +(1.5)/(v)=(1.5-1)/(0.10)=5` `thereforev=(1.5)/(5)=0.3m`from `p_(1)` The IMAGE will serve as a virtual object for II surface. Distance of I. from `P_(2)=(0.3-0.2)=0.10m` `therefore (1.5)/(-0.10)+(1)/(v)=(1.5-1)/(0.10)=5` `(1)/(v) - =5+15=20` or`v=(1)/(20)=0.05m ` from `P_(2)` `therefore ` The final image will be at 0.05 m from `P_(2)`. |
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