# Optics doubt

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1. Dec 16, 2014

### AdityaDev

1. The problem statement, all variables and given/known data
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 u0 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

2. Relevant equations
$$\Delta x=n \lambda$$
$$y=n \lambda /d$$

3. 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 y1 and y2 once the slowest ray comes out of slab,
$$t=l/v => t = l/ ( \mu_0 + d \delta \mu)$$
Hence $$\Delta x = l(\mu_0 + d \delta \mu) \frac {(y_2-y_1) \delta \mu}{(\mu_0 + y_1 \delta \mu)(\mu_0 c + y_2 \delta \mu)}$$

2. Dec 17, 2014

### haruspex

I'm not sure what you have calculated. What is Δx? The difference in the two path lengths?
I feel it is more useful to describe how the two light beams will be changed by the slab. Will they be bent through the same angle?
(It might help to recognise that a rectangular slab with uniformly changing index is equivalent to a more familiar piece of optics equipment.)

3. Dec 17, 2014

### AdityaDev

Close the thread. I have solved the problem.

4. Dec 17, 2014

### AdityaDev

I have solved the problem. The question is of high level. Thank you for the response.
Actually $$\mu_1 = \mu_0 + \delta \mu$$
Find the time taken by slowest ray to come out of slab. Let it be t.
$$t=l \mu_1/c$$
During that time find the distance traveled by fastest ray(at bottom.
Now the situation is like fraunhoffer single slit diffraction.

Last edited: Dec 17, 2014
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