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A glass slab is placed between an object...

A glass slab is placed between an object (O) and an observer (E) with its refracting surfaces AB and CD perpendicular to the line OE. The refractive index of the glass slab changes with distance (y) from the face AB as `mu = mu_(0) (1 + y)`. Thickness of the slab is t. Find how much closer (compared to original distance) the object appears to the observer. Consider near normal incidence only.

Text Solution

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The correct Answer is:
`t - (1)/(mu_(0)) ln (1+t)`
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