Question:
In a Young's double slit experiment, light of $500 \mathrm{~nm}$ is used to produce an interference pattern. When the distance between the slits is $0.05 \mathrm{~mm}$, the angular width (in degree) of the fringes formed on the distance screen is close to :
Correct Option: , 2
Solution:
(2) Given : Wavelength of light, $\lambda=500 \mathrm{~nm}$
Distance between the slits, $d=0.05 \mathrm{~mm}$
Angular width of the fringe formed,
$\theta=\frac{\lambda}{d}=\frac{500 \times 10^{-9}}{0.05 \times 10^{-3}}=0.01 \mathrm{rad}=0.57^{\circ} .$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.