In a Young’s double slit experiment, the intensity at a point is (1/4)^{th} of the maximum intensity,

Question

In a Young's double slit experiment, the intensity at a point is \(\left(\frac{1}{4}\right)^{\text {th }}\)of the maximum intensity, the minimum distance of the point from the central maximum is _______\(\mu \mathrm{m}\).

(Given :\( \lambda=600 \mathrm{~nm}, \mathrm{~d}=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m}\)) 

Answer 1

Correct answer is : 200  

\( \mathrm{I}=\mathrm{I}_0 \cos ^2 \frac{\Delta \phi}{2} \)

\( \frac{\mathrm{I}_0}{4}=\mathrm{I}_0 \cos ^2 \frac{\Delta \phi}{2}\) 

\(\Delta \phi=\frac{2 \pi}{3}=\frac{2 \pi}{\lambda}(\Delta \mathrm{n})\) 

\( \frac{\mathrm{dy}}{\mathrm{D}}=\frac{600 \times 10^{-9}}{3} \) 

\(\mathrm{y}=\frac{2 \times 10^{-7} \times 1}{10^{-3}}=2 \times 100 \mu \mathrm{m}\) 

\(=200 \ \mu \mathrm{m}\)