Question

In the Young's double slit experiment using a monochromatic light of wavelength $\lambda$ the path difference (in terms of an integer $n$) corresponding to any point having half the peak intensity is

• A. $(2n+1) \frac{\lambda}{2}$
• B. $(2n+1) \frac{\lambda}{4}$
• C. $(2n+1) \frac{\lambda}{8}$
• D. $(2n+1) \frac{\lambda}{16}$

Answer 1

Answer: (B)

$\frac{I_{\max }}{2}=I_{ m } \cos ^{2}\left(\frac{\phi}{2}\right)$
$\Rightarrow \cos \left(\frac{\phi}{2}\right)=\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{\phi}{2}=\frac{\pi}{4}$
$\Rightarrow \phi=\frac{\pi}{2}(2 n +1)$
$\Rightarrow \Delta x =\frac{\lambda}{2 \pi} \phi=\frac{\lambda}{2 \pi} \times \frac{\pi}{2}(2 n +1)=\frac{\lambda}{4}(2 n +1)$