Question

Light of wavelength '$\lambda$' is incident on a single slit of width 'a' and the distance between slit and screen is 'D'. In diffraction pattern, if slit width is equal to the width of the central maximum then D =

• A. $\frac{a^2}{\lambda}$
• B. $\frac{a}{\lambda}$
• C. $\frac{a^2}{2\lambda}$
• D. $\frac{a}{2\lambda}$

Hint:

Central maximum width scales inversely with slit size ($W \propto 1/a$). Equating the two yields a quick cross-multiplication step ($a^2 = 2\lambda D$) that isolates $D$ in a single movement on your scratch paper!

Answer 1

The Correct Option is C

Step 1: Recall the central width.
In single-slit diffraction the central bright band has width $W=\frac{2\lambda D}{a}$.

Step 2: Apply the condition.
We are told the slit width equals this width, so $a=\frac{2\lambda D}{a}$.

Step 3: Solve for $D$.
$a^2=2\lambda D$, so $D=\frac{a^2}{2\lambda}$. \[ \boxed{\frac{a^2}{2\lambda}} \]