Question

You illuminate a slit of width 0.0675 mm with light of wavelength 711 nm and observe the resulting diffraction pattern on a screen that is situated 2.21 m from the slit. what is the width, in centimeters, of the pattern\'s central maximum?

Answer 1

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y= \frac{n \lambda D}{a}$
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem, 
$a=0.0675 mm=6.75 \cdot 10^{-5} m$
$\lambda=711 nm = 7.11 \cdot 10^{-7} m$
$D=2.21 m$

And the distance of the first minimum (n=1) from the center of the pattern is
$y= \frac{n \lambda D}{a}= \frac{(1) (7.11 \cdot 10^{-7} m)(2.21 m)}{6.75 \cdot 10^{-5} m}=2.33 \cdot 10^{-2} m =2.33 cm$

The problem asks for the width of the pattern's central maximum. This will be equal to the distance between the first minimum on one side and the first minimum on the other side, so it will be equal to twice the distance we just found:
$\Delta y_{max} = 2 y = 2\cdot 2.33 cm= 4.66 cm$