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in Wave Optic by (26.7k points)
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Allow to lights rays to incident on a screen after passing through two slits

1. Why light is passed through two slits.

2. Find the expression for fringe width

3. What happens to the pattern on the screen when the whole apparatus is dipped in water

1 Answer

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1. Because light has wave nature. Two slits gives effect of coherent sources.

2. Expression for band width:

S1 and S2 are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S1 and S2 .

Hence the path difference, S1O – S2O = 0 So at ‘O’ maximum brightness is obtained. Let ‘P’ be the position of nth bright band at a distance xn from O. Draw S1A and S2B as shown in figure.

From the right angle ∆S1AP

we get , S1P2 = S1A2 + AP2
S1P2 = D2 + (Xn – d/2)2
= D2 + Xn2 – Xnd + \(\frac{d^4}{4}\)

Similarly from ∆S2BP we get,
S2P2 = S2B2 + BP2
S2P2 = D2 + (Xn + d/2)2

= D2 + \(\Big(X^2_n+X_nd+\frac{d^2}{4}\Big)\)

∴ S2P2-S1P2 =

D2+\(\Big(X_n^2+X_nd+\frac{d^2}{4}\Big)\)

\(\Big[D^2+\Big(X_n^2-X_nd+\frac{d^2}{4}\Big)\Big]\)

S2P2 – S1P2 =2xnd (S2P + S1P)(S2P – S1P) = 2xnd

But S1P ≈ S2P ≈ D

∴ 2D(S2P – S1P) = 2xnd

i.e., path difference S2P – S1P = \(\frac{x_nd}{D}\) ......(1)

But we know constructive interference takes place at P, So we can take

(S2P – S1P) = nλ

Hence eq(1) can be written as

\(nλ=\frac{x_nd}{D}\)

\(x_n=\frac{nλd}{D}\)   ......(2)

Let xn+1 be the distance of (n+1)th bright band from centre o, then we can write

\(X_{n+1}=\frac{(n+1)λD}{d}\) .....(3)

∴ band width, b

\(=\frac{(n+1)λD}{d}-\frac{nλD}{d}=\frac{λD}{d}(n+1-n)\)

β = \(\frac{λD}{d}\)

This is the width of the bright band. It is the same for the dark band also.

3. The pattern shrinks (Band width decreases) if whole apparatus is dipped in water. (Because of high refractive index, velocity of light decreases. The wavelength also decreases and hence fringe width also reduce).

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