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

The linear width of the central maximum ($W$) in a standard single-slit Fraunhofer diffraction setup is given by the formula: $$W = \frac{2\lambda D}{a}$$ According to the given problem statement condition, the slit width ($a$) is exactly equal to the linear width of this central maximum ($W$): $$a = \frac{2\lambda D}{a}$$ Rearranging the expression to isolate the screen distance parameter $D$: $$a^2 = 2\lambda D \implies D = \frac{a^2}{2\lambda}$$
Final Answer:
The distance $D$ is equal to $\frac{a^2}{2\lambda}$, which corresponds to option (C).