Question

In Young's double slit experiment, the distance $d$ between the slits $S_1$ and $S_2$ is $0.1\text{mm}$. What should be the maximum width of each slit, so as to obtain ${10}^{th}$ maxima of the double slit pattern within the central maximum of the single slit pattern?

• A. $0.09\text{mm}$
• B. $0.08\text{mm}$
• C. $0.02\text{mm}$
• D. $0.06\text{mm}$

Answer 1

The correct option is C $0.02\text{mm}$

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

$\because x_n=\frac{n\lambda D}{d}$

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

The angular width of the central maximum in the diffraction pattern due to a single slit of width $a$, is,
$2{\theta }_1=\frac{2\lambda }{a}$

Now, $\frac{10\lambda }{d}\le \frac{2\lambda }{a}$

$\frac{5}{d}\le \frac{1}{a}$

$\Rightarrow a\le \frac{d}{5}$

As, $d=0.1\text{mm}$

$\therefore a\le 0.02\text{mm}$

Hence, $\left(C\right)$ is the correct answer.