Question 184: Wave Optics

Question

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm . If the 20 maxima of the double slit pattern are contained within the central maximum of the single slit diffraction pattern, then the width of each slit is $x 	imes 10^{-3} \mathrm{~cm}$, where $x$-value is _________ .

Accepted Answer

Width of 20 maxima of double slit $=$ width of central maxima of single slit
$$\begin{aligned}
& \frac{20 \lambda \mathrm{D}}{\mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \
& \frac{10}{\mathrm{~d}}=\frac{1}{\mathrm{a}} \
& \mathrm{a}=\frac{\mathrm{d}}{10}=\frac{1.5 	imes 10^{-1}}{10} \mathrm{~cm}=15 	imes 10^{-3} \mathrm{~cm}
\end{aligned}$$
Value of $x$ is 15
Answer is 15

Wave Optics

Integer / Numeric Answer Type