In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern.
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 :

  1. $0.17^{\circ}$

  2. $0.57^{\circ}$

  3. $1.7^{\circ}$

  4. $0.07^{\circ}$


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,

$\theta=\frac{\lambda}{d}=\frac{500 \times 10^{-9}}{0.05 \times 10^{-3}}=0.01 \mathrm{rad}=0.57^{\circ} .$

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