Question

I need B and C

Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0240 mm wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.90 W/m² Part A Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. Express your answer as an integer. mmax = 74 VUILLING Part B At what angle does the dark fringe that is most distant from the center occur? Express your answer in degrees. IVO AEO ? 10max| = 88.95 Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Part C What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it. | ΑΣφ ? I = 3.61 W/m2

Answer 1

Solution:

Given data

The wavelength of the incident light,

X= 632.8 nm

Width of the slit,

d = 0.0240 mm

The intensity at the center of the central bright fringe,

1. = 8.90 W/m

(9)

The maximum order of diffraction can be calculated as follows

sin ma d

$\frac{m\lambda }{d}=1\; \; \; \; \; \; \; \; \; (\sin \theta =1)$

$m=\frac{d}{\lambda }$

$m=\frac{0.024\times 10^{-3}}{632.8\times 10^{9} }$

$m=37.9\approx 37$

Now, we calculate the angle corresponding to this order as follows

sin ma d

$\theta =\sin ^{-1}\left ( \frac{m\lambda }{d} \right )$

$\theta_{37} =\sin ^{-1}\left ( \frac{37\times 632.8\times 10^{-9}}{0.024\times 10^{-3}} \right )$

$\theta_{37} =77.42^{\circ}$

Now, calculate the angle corresponding to m = 36

$\theta_{36} =\sin ^{-1}\left ( \frac{36\times 632.8\times 10^{-9}}{0.024\times 10^{-3}} \right )$

$\theta_{36} =71.62^{\circ}$

Center of the fringe can be calculated as follows

$\theta_{max} =\frac{77.42+71.62}{2}$

$\theta_{max} =74.52^{\circ}$

(c)

The intensity of the bright fringe before the dark fringe has seen in part (b) can be calculated as follows

$\beta =\frac{2\pi }{\lambda }d\sin \theta$

$\beta =\frac{2\times 3.14}{632.8\times 10^{-9} }(0.024\times 10^{-3})\sin 74.52$

$\beta =229$

Now, the intensity of the bright fringe can be calculated as follows

$I=I_{o}\frac{(\sin \frac{\beta }{2})^{2}}{(\frac{\beta }{2})^{2}}$

$I=I_{o}\frac{(\sin \frac{229 }{2})^{2}}{(\frac{229 }{2})^{2}}$

$I=I_{o}\times 6.32\times 10^{-5}$

We know,

1. = 8.90 W/m

Then we have

$I=8.90\times 6.32\times 10^{-5}$

$I=56.25\times 10^{-5}$

$I=5.625\times 10^{-4}\; \mathrm{W/m^{2}}$