Question

Obtain the expression for fringe width in the case of interference of light waves.

Answer 1

Light rays coming from slits $S_1$ and $S_2$ interfere at point P.

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

Or $S_{1}P=\sqrt{y^2+\frac{d^2}{4}-yd+D^2}$

Or $S_{1}P=D\sqrt{\frac{y^2}{D^2}+\frac{d^2}{4D}-\frac{yd}{D^2}+1}$

As $D>>d$ and $y<<D$, we neglect $d^2/4D$ and $y^2/D^2$.

We get

Or $S_{1}P=D\sqrt{1-\frac{yd}{D^2}}$

Or $S_{1}P=D[1-\frac{yd}{2D^2}]$

Or $S_{1}P=D-\frac{yd}{2D}$

Similarly we get $S_{2}P=D+\frac{yd}{2D}$

Path difference between two light rays $\Delta x=S_{2}P-S_{1}P$

Or $\Delta x=(D+\frac{yd}{2D})-(D-\frac{yd}{2D})=\frac{yd}{D}$

For constructive interference, $\Delta x=n\lambda$

$\therefore$ $\frac{yd}{D}=n\lambda$

$⟹$ $y=\frac{n\lambda D}{d}$

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

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

$⟹$ $\beta =\frac{\lambda D}{d}$