PHYSICS 41Physics 41 Chapter 38 HW Key question and answers

Physics 41 Chapter 38 HW Key

1. Helium–neon laser light (632..8 nm) is sent through a 0.300-mm-wide single slit. What is the width
of the central maximum on a screen 1.00 m from the slit?

λ 6.328 × 10−7 y
sin θ= = = 2.11× 10−3 = tan θ ≈ sin θ ≈ θ (for small θ)
a 3.00 × 10−4 1.00 m
2y = 4.22 mm


2. A beam of monochromatic green light is diffracted by a slit of width 0.550 mm. The diffraction
pattern forms on a wall 2.06 m beyond the slit. The distance between the positions of zero intensity
on both sides of the central bright fringe is 4.10 mm. Calculate the wavelength of the light.
y λ
P38.2 The positions of the first-order minima are ≈ sin θ =
± . Thus, the spacing between these two minima is
L a
λ
∆y =2   L and the wavelength is
 a
−3 −3
 ∆y  a   4.10 × 10 m  0.550 × 10 m 
=λ =     =  547 nm
 2  L   2  2.06 m 

Coherent light of wavelength 501.5 nm is sent through two parallel slits in a large, flat wall. Each slit is
0.700 µ m wide. Their centers are 2.80 µ m apart. The light then falls on a semicylindrical screen, with
its axis at the midline between the slits. (a) Predict the direction of each interference maximum on the
screen as an angle away from the bisector of the line joining the slits. (b) Describe the pattern of light
on the screen, specifying the number of bright fringes and the location of each. (c) Find the intensity of
light on the screen at the center of each bright fringe, expressed as a fraction of the light intensity Imax
at the center of the pattern.

Solution:
First some review:

The total intensity of the double slit interference
pattern is a combination of single slit diffraction and
the double slit interference. The cosine squared term
gives the intensity of the fringes inside the single slit
maxima. The sine squared terms is the ‘envelope’ that
shapes the fringes.


 πd sin θ   sin (πa sin θ / λ ) 
2

I = Imax cos 
2
 
 λ   πa sin θ / λ 



Without the diffraction envelope all of the fringes would have equal intensity. This is what we were
shown in chapter 37, although this pattern doesn’t really exist.

, So for this problem, you use the full expression to find the intensity of the bright fringes.
The single slit minima DOMINATE the double slit interference so if a double maxima falls
on a single slit minima, the spot will be dark.

To find the location of the fringes:
Double-slit interference maxima are at angles given by d sin θ = mλ .
Single-slit diffraction minima, at asin θ = mλ .


(a) The ‘semicylindrical screen’ is irrelevant. The point is, they include that because it would be difficult to SEE
fringes at large angles unless the screen was curved – but it doesn’t matter when you are solving for the
angles.

Double-slit interference maxima are at angles given by d sin θ = mλ .

For m = 0 , θ 0= 0°

For m = 1 , ( 2.80 µ m ) sin θ = 1( 0.501 5 µ m ) =
: θ1 sin −1 ( 0.179
= ) 10.3°

Similarly, for m = 2, 3, 4, 5 and 6, θ2
= , θ3
21.0° = , θ4
32.5° = 45.8°

θ5
= , and θ 6 sin
63.6°= = −1
( 1.07 ) nonexistent

Thus, counting the central fringe, there are 5 + 5 + 1 =11 directions for interference maxima .

(b) We check for missing orders by looking for single-slit diffraction minima, at asin θ = mλ because
the single slit diffraction DOMINATES the double slit interference:
For m = 1 , ( 0.700 µ m ) sin θ = 1( 0.501 5 µ m ) and θ1 45.8°
=

Thus, there is no bright fringe at this angle. There are only nine bright fringes , at
θ = 0°, ± 10.3°, ± 21.0°, ± 32.5°, and ± 63.6° .

c)



To find the ratio of intensities, you have to use this:

 πd sin θ   sin (πa sin θ / λ ) 
2

I = Imax cos  2
 
 λ   πa sin θ / λ 