Question

3.Monochromatic green light of wavelength 550nmilluminates two parallel narrow slits 7.7micro meter apart.The angular deviation "theeta" of third order (for m=3) bright fringe a) In radian and b) In degree --------------

Answer 1

We know the relationship for bright fringe is given by
$d\sin \theta =m\lambda$
Now d is the separation between the slit $7.7\text{μm}=7.7\times {10}^{-6}\text{m}$, m is the order of interference = 3 and $\lambda$ is the wavelength of the light $=550\text{nm}=550\times {10}^{-9}\text{m}$. So the angle of diversion is given by
$\theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)={\sin }^{-1}\left(\frac{3\times 550\times {10}^{-9}}{7.7\times {10}^{-6}}\right)={\sin }^{-1}\left(0.21\right)=0.21rad=12.03°$

1) The angular deviation in radian is 0.21rad
2) Angular deviation in degree is 12.03^o