Question

In a young's double slit experiment $\lambda=500 nm , d =$ $1.0 mm$ and $D =1 m$. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity:

• A. $1.25 \times 10^{-3} m$
• B. $1.25 \times 10^{-4} m$
• C. $2.25 \times 10^{-1} m$
• D. $125 m$

Answer 1

Answer: (B)

$I=4 I_{0} \cos ^{2} \phi / 2$
$\frac{\left(4 I_{o}\right)}{2}=4 I_{o} \cos ^{2} \phi / 2$ $\quad$ [Intensity is half of the maximum]
$\therefore \frac{1}{\sqrt{2}}=\cos \phi / 2$
$90^{\circ}=\phi$
Minimum distance from central maximum corresponding to $\phi=90^{\circ}$ is
$x =\frac{\phi}{2 \pi} \times \beta$
$\Rightarrow \frac{\pi / 2}{2 \pi} \times \frac{\lambda D }{ d }$
$x =\frac{1}{4} \times \frac{500 \times 10^{-9} \times 1}{1 \times 10^{-3}} $
$x =1.25 \times 10^{-4} m$