Question

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

Answer 1

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 n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.

From the right angle ∆S₁AP

we get , S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² – X_nd + \(\frac{d^4}{4}\)

Similarly from ∆S₂BP we get,
S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + 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 =2x_nd (S₂P + S₁P)(S₂P – S₁P) = 2x_nd

But S₁P ≈ S₂P ≈ D

∴ 2D(S₂P – S₁P) = 2x_nd

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 x_n+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).