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## Homework Statement

A rectangular slab of length l=20cm and thickness d=4cm is placed in left of a converging lens of focal length f=20cm. A screen is placed in the focal plane plane of lens (right side of lens).Refractive index of the material of slab increases linearly from u

_{0}at the bottom by an amount ##du=2*10

^{-4})## between bottom and top faces.A parallel beam is made to incident on the left side of slab. Prove that only a bright spot forms on the screen? Find its position

## Homework Equations

[tex]\Delta x=n \lambda[/tex]

[tex]y=n \lambda /d[/tex]

## The Attempt at a Solution

Here as the refractive index varies, speed of light varies. It will be faster at bottom and slower at top.

$$v=c/ \mu$$

$$ \mu = \mu_o + y \delta \mu $$

For any two rays at y

_{1}and y

_{2}once the slowest ray comes out of slab,

$$t=l/v => t = l/ ( \mu_0 + d \delta \mu)$$

Hence $$ \Delta x = l(\mu_0 + d \delta \mu) \frac {(y_2-y_1) \delta \mu}{(\mu_0 + y_1 \delta \mu)(\mu_0 c + y_2 \delta \mu)} $$