Question

A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the first two minima on either side of the central maximum?

Answer 1

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

$a\sin \theta_n=n\lambda$

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

$a\sin \theta_1=(1)\lambda$

solving for θ1:

$\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_1=0.63 degrees$

for the second minimum:

$a\sin \theta_2=(2)\lambda$

$\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_2=1.26 degrees$

So, the angular separation between them is the rest:

$\Delta \theta =1.26-0.63$

$\Delta \theta=0.63$