Double Slit Diffraction Problems
Answers
Young’s Double Slit Formula: m  d sin(  )
h
d
θ
l
1) In a double slit interference experiment, the distance between the slits is 0.0005m and
the screen is 2 meters from the slits. Yellow light from a sodium lamp is used and it has a
wavelength of 5.89 x 10-7 m. Show that the distance between the first and second fringes
on the screen is 0.00233 m. (Fringe is another word for bright spot).
 =5.89 x 10-7 m
d= 5.0 x10 4 m
d sin   m  sin   m / d    sin 1 m / d 
m=1, 1 = 0.675º
m=2.  2 = 0.135 º
tan   h / l
y1  h
tan   y1 / l  y1  l tan   2  tan 0.675
y1 (Height of lower fringe) = 0.0236m
y 2 (Height of upper fringe) = 0.0472m
y  .0236m
y Difference between two fringe locations. (For small angles – approximately evenly
spaced – You could take distance to third fringe and divide by 3).
2) With two slits spaced 0.2mm apart, and a screen at a distance of l=1m, the third bright
fringe is found to be displaced h=7.5mm from the central fringe. Show that the
wavelength,  , of the light used is 5  10 7 m .
Step 1, Given l, h, find the angle  by using trigonometry.
tan   h / l    tan 1 h / l   0.430
Step 2. Now that  is known, you can use the formula with m=3 to find the wavelength.
d sin   m    d sin  / m  2.0 x104  sin 0.430 / 3  5.00 x107
3. Two radio towers are broadcasting on the same frequency. The signal is strong at A,
and B is the first signal minimum. If d = 6.8 km, L = 11.2 km, and y = 1.73 km, what is
the wavelength of the radio waves to the nearest meter?
First find the angle:
tan   y / L    tan 1  y / L   tan 1 1.73 / 11.2  8.78
Find the wavelength
d sin   m    d sin  / m  6.8  sin 8.78 / 1  1.04km
Double Slit Diffraction Problems
Answers
Young’s Double Slit Formula: m  d sin(  )
4a. Water waves of wavelength of 5.44 meters are incident upon a breakwater with two
narrow openings separated by a distance 247 meters. To the nearest thousandth of a
degree what is angle corresponding to the first wave fringe maximum?
d  247m
  5.44m
m  d sin   sin   m / d    sin 1 m / d 
  sin 1 1 5.44 / 247   1.262
5. In a double-slit experiment it is found that blue light of wavelength 467 nm gives a
second-order maximum at a certain location on the screen. What wavelength of visible
light would have a minimum at the same location?
Two different experiments with the same slit spacing, d, and same angle, θ.
m11  m2  12 2
2  467nm  m2  12 2
If m2  0,  2  1868nm This is not visibl e light.
If m2  1,  2  622nm. This is visible light
If m2  2,  2  374nm This is visible light.
If m2  3,  2  267nm. This is not visibl e light
6. Find the distance between adjacent dark spots from a double slit diffraction pattern if
the wavelength of light is 500 nm, the distance between the slits is 1 mm, and the
distance from the slit to the screen is 2 m.
Since spots are almost uniformly spaced, the distance between dark spots is the same as
the distance between bright spots. So just find distance to first bright spot.
-First find the angle:
d sin   m  sin   m / d 
  sin 1 m / d   sin 1 1 5 x107 / 1.0 x103 
  0.0286
-Now use that angle to find the height of the first bright spot
tan   h / l  y1  l tan   2m  tan( 0.0286)  0.00100m