Question

A single slit of width 0.50 mm is illuminated with light of wavelength 500 nm, and a screen is placed 120 cm in front of the slit. Find the widths (lengths, not angles)of(a) the central maximum,(b) the first maximum beyond the central maximum, and(c) the second maximum beyond the central maximum.

Answer 1

Answer:

a) 2.4 mm

b) 1.2 mm

c) 1.2 mm

Explanation:

To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:

$m\lambda=asin\theta$

a: width of the slit

λ: wavelength

m: order of the minimum

for little angles you have:

$y=\frac{m\lambda D}{a}$

y: height of the mth minimum

a) the width of the central maximum is 2*y for m=1:

$w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm$

b) the width of first maximum is y2-y1:

$w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm$

c) and for the second maximum:

$w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm$