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Location of image after multiple events Find the location of final image as shown in Fig.34-13. Take the refractive index of the slab as 4/3. |
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Answer» Solution :(1) This is again an example of multiple events before final image is FORMED. The rays coming from O will pass through the slab, get refracted by it and then reflected by the mirror, and again refracted by the slab. This refraction will from the final image. Whenever we refer to the image of the system, it refers to this image. (2) The mirror will not see the object at 15cm but shifted toward it. After reflection, image will not be directly formed at the point where the mirror focuses the light but will be shifted in the direction of light incident on the slab. Calculation : The very first event is the refraction from the glass slab. It causes a SHIFT in the direction of the incident rays. This normal shift is given by Eq. 34-8. `x=t(1-(1)/(n))` `=4xx(1-(3)/(4))=1cm` Thus, the object will seem to be shifted toward the mirror (in the direction of the incident rays). So, the mirror will see the object at `15-1=14cm`. So, for the mirror `u=-14cm` . After reflection from the mirror, `v=(14xx10)/(14-10)=-35cm` One more time reflection will occur as light after reflecting from mirror will pass through the slab. Again, normal shift will be 1cm, but in the direction of propagation of light. Now the light rays are propagating in the LEFT direction. Therefore, final image will be at a DISTANCE of `36cm` from the mirror. Learn : If there are more than one slabs, the total normal shift as Normal shift =`t_(1)[1-(n)/(n_(1))]+t_(2)[1-(n)/(n_(2))]+......` 
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