Question

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm.
If the intensity at the center of the central maximum is 5.0010?4W/m2 , what is the intensity at a point on the screen that is 0.830mm from the center of the central maximum?

Answer 1

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength,$\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm=$\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$