# Diffraction and interference

Given: a=44μm and L= 1.5m

1. Let the wavelength of the light beλ=443nmWhat is the distance on the screen between the most intense maximum and the nearest minimum in the pattern of emerging light, in mm?

2. Now the single slit is replaced by a pair of very narrow parallel slits separated by d=32μm. A laser with a wavelength λ=420nm is used. What is the distance on the screen between two neighboring maximum in the emerging light patter in mm?

Answer

1) the diffraction takes place for single slit between an ending of the slit and its middle.

$d = a/2 =22*10^{-6} (m)$

maximum diffraction for

$d*sin(\theta) =k*\lambda$

minimum diffraction for

$d*sin(\theta) = (k+1/2)*\lambda$

$sin(\theta) = tan(\theta) = D/L$

most intense maximum is $k =0$, means $D(max) =0$ m (it is the center fringe on the screen)

next minimum is for

$d*D/L =1/2*\lambda$

$D = 1/(2d)*\lambda*L =1/(2*22*10^{-6})*443*10^{-9}*1.5 =0.0151 m =1.51 cm =15.102 mm$

2) the diffraction takes place between the two openings of the two slits.

maximum condition is

$d*sin(\theta) =k*\lambda$

$sin(\theta) =tan(\theta) = D/L$

$d*D/L =k*\lambda$

$k_1 =0$ and $k_2=1$

$D =\lambda*L/d = 420*10^-9*1.5/32*10^-6 =19.6875 mm$