Question

In a single-slit diffraction experiment, the width of the slit is reduced by half. Which of the following needs to be done, so that, the width of the central maxima remains the same.

• A. Reduce the distance between the slit and screen by half
• B. Reduce the distance between the slit and the screen to ${\left(\frac{1}{4}\right)}^{th}$ the original separation.
• C. Double the distance between the slit and the screen.
• D. No need to do anything, as the width of the central maxima does not depend on the slit width

Answer 1

The correct option is A Reduce the distance between the slit and screen by half

Width of central maximum is, $=2\beta =\frac{2\lambda D}{a}$
Where $a$ is the width of the slit and $D$ is the distance between the screen and slit.

$\Rightarrow \beta \propto D$ and $\beta \propto \frac{1}{a}$

When the width of the slit is reduced by half, the width of central maximum is doubled. So, in order to keep the width of the central maximum unchanged, we have to reduce the distance between the screen and slit by half.
$\therefore$ New width of central maximum is, $=2{\beta }^{\prime }=\frac{2\lambda \left(\frac{D}{2}\right)}{\left(\frac{a}{2}\right)}=\frac{2\lambda D}{a}=2\beta$

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

Why this question? To understand the factors affecting the width of central maximum.