Question

You illuminate a slit with a width of 71.7 μm with a light of wavelength 737 nm and observe the resulting diffraction pattern on a screen that is situated 2.27 m from the slit. What is the width, in centimeters, of the pattern's central maximum?

Answer 1

To solve this problem we will apply the concepts related to the double slit-experiment. Under this concept we understand the relationship between the minimum angle, depending on the order of the fringes, the wavelength and the distance between slits. Therefore we have the following relation,

$sin(\theta_{min}) = \frac{m\lambda}{D}$

Here,

m = Order of the fringes

D = Distance between slits

$\lambda$ = Wavelength

Replacing with our values we have,

$sin(\theta_{min}) = \frac{(1)(737*10^{-9})}{71.7*10^{-6}m}$

$sin(\theta_{min}) = 0.01028$

Through the relationship between distances then we have that the basic amplification distance is given by the relationship between the distance of the slit L and the angle, then

$Y=Lsin\theta$

$Y =2.27*0.01028$

$Y =0.0233m$

Thus the width of the central maximum is

$W=2Y=2*0.0233$

$W = 0.0466m$

Therefore the widht is 0.466m