1. Two narrow slits 46 µm apart are illuminated with light of wavelength 550 nm.
ID: 2309852 • Letter: 1
Question
1.
Two narrow slits 46 µm apart are illuminated with light of wavelength 550 nm.
What is the angle of the m = 2 bright fringe in radians?
What is the angle in degrees?
2.
A pair of slits is located near a screen in such a way that light incident on the slits can form a double slit interference pattern on the screen.
(a) Which of the following actions performed by itself would increase the angular position of the m = 3 maximum of the pattern (more than one answer may be correct)?
use light with a longer wavelengthuse light with a shorter wavelengthincrease the distance from the slits to the screendecrease the distance from the slits to the screenuse slits which are farther apartuse slits which are closer together
(b) Which of the following actions performed by itself would decrease the distance along the screen between the pattern's center and the m = 5 minimum of the pattern (more than one answer may be correct)?
use light with a longer wavelengthuse light with a shorter wavelengthincrease the distance from the slits to the screendecrease the distance from the slits to the screenuse slits which are farther apartuse slits which are closer together
3.
A double-slit interference pattern is created by two narrow slits spaced 0.18 mm apart. The distance between the first and the fifth minimum on a screen 70 cm behind the slits is 6 mm. What is the wavelength of the light used in this experiment?
4.
If sunlight shines straight onto a peacock feather, the feather appears blue when viewed from 23° on either side of the incident beam of sunlight. The blue color is due to diffraction from the melanin bands in the feather barbules. Blue light with a wavelength of 467 nm is diffracted at 23° by these bands (this is the first-order diffraction) while other wavelengths in the sunlight are diffracted at different angles. What is the spacing of the melanin bands in the feather?
Explanation / Answer
1) d = 46 mum = 46 * 10-6 m
lambda = 550 nm = 550 * 10-9 m
d sin theta = m lambda
sin theta = 2 lambda / d
sin theta = (2 * 550 * 10-9 ) / (46 * 10-6)
theta = 1.37 deg
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3) for first minimum m = 0
for fifth minimum m = 4
delta y = y4 - y1 = [(lambda L / d) * (4 + 1/2)] - [(lambda L / d) * (0 + 1/2)]
delta y = 4 lambda * L / d
lambda = d * delta y / 4 L
= (0.18 mm * 6 mm) / (4 * 70 cm)
= 386 nm
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