In a double-slit interference experiment the slit separation is 8.40 x 10-6 m an

Question

In a double-slit interference experiment the slit separation is 8.40 x 10-6 m and the slits are 2.80 m from the screen. Each slit has a width of 1.20 x 10-6 m. a) An interference pattern is formed when light with a wavelength of 450 nm is shined on the slits. How far (in meters) from the center of the interference pattern on the screen do the third order (m = 3) bright fringes occur? b) If a second light source with a wavelength of 600 nm is shined on the same slits, how far (in meters) from the center of the interference pattern on the screen will the third order (m = 3) dark fringes occur for this wavelength? c) If both of the above light sources are shined on the slits at the same time, which order bright fringes from the two light sources would occur at the same locations on the screen? Include the orders of all possible fringes that would overlap. d) Describe qualitatively how the pattern on the screen would change if one of the two slits is covered up so that you now have single-slit diffraction. e) What is the total angular width of the central diffraction maximum for the 450 nm light source when one slit is covered up? f) How many bright fringes from the double-slit pattern created by the 450 nm light source (part a) ) would fit within the central diffraction maximum created when one slit is covered? (part e) )?

Explanation / Answer

a)
y = 3*lambda*D/d
= 3* (450*10^-9)*2.8 / (8.4*10^-6)
= 0.45 m
Answer: 0.45 m

b)
y = 3*lambda*D/d
= 3* (600*10^-9)*2.8 / (8.4*10^-6)
= 0.6 m
Answer: 0.6 m

c)
4th order of 450 nm will coincide with 3 rd order of 600 nm
8th order of 450 nm will coincide with 6th order of 600 nm

and so on.

d)
if one slit is covered, we will get diffraction pattern.
here central maxima will be double the width of others maxima

e)
angular width = 2*lambda/d
= 2*450*10^-9 / (1.2*10^-6)
=0.75 rad
Answer: 0.75 rad

I am allowed to answer only 4 parts at a time

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