MCQ
In Young's double slit experiment, at a fixed point where $\lambda$ is path difference $=\lambda / 6(\lambda=$ wavelength of the light used), the intensity is I. If $I _0$ is the maximum intensity then $I / I_0$ is equal to :
  • $\frac{3}{4}$
  • B
    $\frac{1}{\sqrt{2}}$
  • C
    $\sqrt{3} / 2$
  • D
    $\frac{1}{2}$

Answer

Correct option: A.
$\frac{3}{4}$
(A) - $\frac{3}{4}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free