InterviewSolution
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InterviewSolution
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Image of an object kept on the principal axis of transparent sphere with silvered rear surface A sphere of radius 1m and n=1.5 is silvered at its back. A point objet is kept at a distance of 1m from the front face (Fig. 34-30). Where will the final image be formed ? |
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Answer» Solution :Here , we can clearly see that there will be three events before the final IMAGE will be formed. Refraction, reflection , and refraction back into the air. Calculation : For the FIRST event O is the optical center, `n_(1)=1` , `n_(2)=1.5`, `u=-1m` , `R=+1m`. Therefore, we get `(1.5)/(v_(1))-(1)/(-1)=(1.5-1)/(1)` `v_(1)=3m` This image will act as an object for the rear silvered surface . However, for TAKING the object distance, we have to MEASURE it from the pole of the mirror. As we have discussed before, `u_(2)=v_(1)-x` `u_(2)=-3-2=-5m` Applying mirror formula, we get (radius of curvature is `-1m`, so final length of the mirror is `-1//2`m) `(1)/(v_(2))+(1)/(-5)=(1)/(-0.5)` `v_(2)=-(5)/(9)m`. These reflected rays will again be refracted from the spherical surface . Since now The direction of the light rays has reversed, therefore `u_(3)=-v_(2)-x` `u_(3)=-(-(5)/(9))-2=-(13)/(9)m`. This can be seen from Fig. 35-29 also. Now, `(1)/(v_(3))-(1.5)/((-13)/(9))=(1-1.5)/(-1)=0.5` Note here that the radius of this surface is negative becuase the direction of the incident rays has reversed. Solving we get `v_(3)=-13//7m`. Negative means that the image is virtual formed inside the spherical surface. This is distance from the left edge of the spherical surface. 
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