1. Light with a wavelength of 589 nm illuminates two narrow slits. A fringe spac
ID: 582227 • Letter: 1
Question
1. Light with a wavelength of 589 nm illuminates two narrow slits. A fringe spacing of 4.0 mm is on a screen 150 cm behind the slits. What is the spacing between the two slits?
2. The second minimum in the diffraction pattern of a 0.10 mm wide slit occurs at 0.7 degrees. What is the wavelength of the light?
3. A double slit experiment is performed with light of wavelength 600 nm. The bright interference fringes are spaced 1.8 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 400 nm?
4. The human eye can see from 400 nm to 700 nm. White light illuminates a diffraction grating having 750 lines/mm. Over what range of angles (in degrees) does the visible m=1 spectrum extend?
Explanation / Answer
1)
fringe width = wavelength*D/d
Spacing between the slits is d = wavelength*D/fringe width
d = 589*10^-9 *150*10^-2/ 4*10^-3
d = 0.22 mm
2)
a=width of slit
m=integer multiple
=wavelength
=angle
asin=m (at m=2; mN)
(0.0001)sin(0.7)=2()
=610.8 nm
3)
fringe width = wavelength*D/d
New fringe width = f1*w2/w1 = 1.8*400/600 = 1.2 mm
4)
750 lines/mm = 750000 lines/m --> the distance is 1/7.5*10^5 m = 1.332*10^-6 m
d*sin =
sin = /d
1 = sin^-1(400*10^-9/(1.332*10^-6)) = sin^-1(0.292) = 17.47°
2 = sin^-1(700*10^-9/(1.332*10^-6) = sin^-1(0.511) = 31.7°
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