Target Exam

CUET

Subject

Physics

Chapter

Wave Optics

Question:

A slit of width e is illuminated by light of wavelength λ. What should be the value of e to obtain the first maximum at an angle of diffraction of $\frac{\pi}{3}$?

Options:

$\frac{2}{\sqrt{3}}λ$

$\frac{λ}{\sqrt{3}}$

$\sqrt{3}λ$

$\frac{\sqrt{3}}{2}λ$

Correct Answer:

$\sqrt{3}λ$

Explanation:

$\text{For single slit diffraction, maxima condition (approx): } e\sin\theta = \frac{(2m+1)\lambda}{2}$

$\text{First maximum: } m=1$

$e\sin\theta = \frac{3\lambda}{2}$

$\theta=\frac{\pi}{3} \Rightarrow \sin\theta=\frac{\sqrt{3}}{2}$

$e \cdot \frac{\sqrt{3}}{2} = \frac{3\lambda}{2}$

$e = \frac{3\lambda}{\sqrt{3}} = \sqrt{3}\lambda$

$e = \sqrt{3}\lambda$