Question

In Young's double slit experiment the distance d between the slits $S_1$ and $S_2$ is 1 mm. What should the width of each slit be so as to obtain ${10}^{th}$ maxima of the double slit pattern within the central maximum of the single slit pattern?

• A. 0.9 mm
• B. 0.8 mm
• C. 0.2 mm
• D. 0.6 mm

Answer 1

The correct option is C 0.2 mm

The angular separation between n bright fringes, ${\theta }_n=\frac{x_n}{D}=\frac{n\lambda }{d}$ $\left[\because x_n=\frac{n\lambda D}{d}\right]$

For 10 bright fringes, ${\theta }_{10}=\frac{10\lambda }{d}$

The angular width of the central maximum in the diffraction pattern due to slit of width a is

$2{\theta }_1=\frac{2\lambda }{a}$

Now $\frac{10\lambda }{d}\le \frac{2\lambda }{a}$ or $a\le \frac{d}{5}=\frac{1}{5}mm$ = 0.2 mm