In a young's double slit experiment, the fringe pattern is observ… - Vidyadip

MCQ

In a young's double slit experiment, the fringe pattern is observed on a screen placed at a distance $D$. The slits are separated by $d$ and are illuminated by light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to half the maximum is

✓
$\frac{\lambda D}{4d}$
B
$\frac{\lambda D}{2d}$
C
$\frac{\lambda D}{d}$
D
$\frac{2}{3} \frac{\lambda D}{d}$

✓

Answer

Correct option: A.

$\frac{\lambda D}{4d}$

a
$I=I_{\max } \cos ^{2} \frac{\phi}{2}$

$\frac{I_{\max }}{2}=I_{\max } \cos ^{2} \frac{\phi}{2} \therefore \cos \frac{\phi}{2}=\frac{1}{\sqrt{2}}$

$\frac{\phi}{2}=\frac{\pi}{4} \quad \phi=\frac{\pi}{2}$

Path diff. $\Delta \mathrm{x}=\left(\frac{\lambda}{2 \pi}\right)(\Delta \phi)=\frac{\lambda}{2 \pi} \times \frac{\pi}{2} \Rightarrow \frac{\lambda}{4}$

Distance $\quad \mathrm{x}=\frac{\Delta \mathrm{D}}{\mathrm{d}}=\frac{\lambda}{4} \frac{\mathrm{D}}{\mathrm{d}}$

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