Question

To produce interference fringe pattern, draw a ray diagram of Young's double slit experiment. Derive an expression of fringe width for bright interference fringes. If fringe width of bright fringes is $2mm$, write the fringe width for dark fringes. Whether the center point of interference fringe pattern is bright or dark? Explain it.

Answer 1

Let the separation between the slits in Young's double slit experiment be $d$ and $\lambda$ be the wave length of light used.

Also, the light reaching at point P via different paths interfere and formed $nth$ bright fringe at point P.

From figure, $S_{1}P=\sqrt{S_1{B}^2+P{B}^2}=\sqrt{D^2+(y_n-\frac{d}{2}{)}^2}$

Similarly, $S_{2}P=\sqrt{S_2{A}^2+P{A}^2}=\sqrt{D^2+(y_n+\frac{d}{2}{)}^2}$

Path difference between the light reaching at P $\Delta x=S_{2}P-S_{1}P$

$\therefore$ $\Delta x=$ $\sqrt{D^2+(y_n+\frac{d}{2}{)}^2}-$ $\sqrt{D^2+(y_n-\frac{d}{2}{)}^2}$

OR $\Delta x=$ $D[\sqrt{1+\frac{(y_n+\frac{d}{2}{)}^2}{D^2}}-$ $\sqrt{1+\frac{(y_n-\frac{d}{2}{)}^2}{D^2}}]$

Using $(1+nx{)}^n\approx 1+nx$ for $x<<1$

For $d<<D$, we get $\Delta x=$ $D[1+\frac{(y_n+\frac{d}{2}{)}^2}{2D^2}]-$ $[1+\frac{(y_n-\frac{d}{2}{)}^2}{2D^2}]$

OR $\Delta x=$ $\frac{1}{2D}\left[\right(y_n+\frac{d}{2}{)}^2-$ $(y_n-\frac{d}{2}{)}^2]$

OR $\Delta x=$ $\frac{1}{2D}\left[\right(y_n^2+\frac{d^2}{4}+y_{n}d{)}^2-$ $(y_n^2+\frac{d^2}{4}{)}^2+y_{n}d]$

$⟹$ $\Delta x=\frac{y_{n}d}{D}$

For bright fringe to be obtained, path difference must be equal to integral value of lambda i.e. $\Delta x=n\lambda$

$\therefore$ $\frac{y_{n}d}{D}=n\lambda$ $⟹y_n=\frac{n\lambda d}{D}$

Fringe width of bright fringe $\beta =y_n-y_{n-1}$

$\therefore$ $\beta =n\frac{\lambda D}{d}-(n-1)\frac{\lambda D}{d}$

$⟹$ $\beta =\frac{\lambda D}{d}$

In Young's double slit experiment, bright and dark fringes of equal width are formed.

Fringe width of bright fringe $\beta =2mm$ (Given)

Thus width of dark fringes ${\beta }^{\prime }=\beta =2mm$

As point O is equally spaced for both the slits, thus path difference between the lights reaching the point O is zero and hence bright fringe is formed at point O. Thus centre point of interference fringe pattern is bright.