Physics Lab Report

Interference and Diffraction
February 4, 2016
Section 002
Gary Schultz and Rachel Weyand

Abstract:
The purpose of this lab was to calculate the wavelength of light intensity for single and
double slits. When the light passes through single slits, the result is interference and diffraction.
For purposes of this lab, the interference wave was used by experimentally measuring between
the first minima on either side of the centroid for values of a. Intensity was found through
where Im is the maximum intensity and is given by Here, a is(sinα/α)I = Im
2
α πasinθ/λ). α = (
the width of each slit, already measured, and is the wavelength. The slope of the best fit curveλ
of sin( vs 1/a was used to find wavelength. The manufacturer’s is 670 nm and the)θ λ 0 ± 1
experimental is 600 10 nm. This yielded a 10.4% error. For both experiments, the method ofλ ±
measuring the distance between the minima created a significant amount of error which is
discussed in the conclusion.
Experiment two of the lab used a double slit instead and the diffraction wave was used by
measuring the distance between the first four minima on either side of the central maximum
using the same laser. For intensity of a double slit, the equation is , where(cosβ) (sinα/α)I = Im
2 2

is the same as the first experiment and is given by . The distance between theα β πdsinθ/λ)β = (
slits is given by d and is provided. 620 20 nm was the experimental wavelength calculated±
using averages. The percent error for experiment 2 was 7.46%.

Data:
Table 1: The table below shows the measured lengths of the central maximum created by single
slits of various lengths. Here,  x is the measured distance between minima.
Data number width of slit (a)
(mm)
x1 (mm) x2 (mm) average x (mm)
1 0.02 62.0 .05± 0 59.9 0.05± 60.95 0.7±
2 0.04 34.0 0.03± 29.8 0.03± 31.9 1.0±
3 0.08 15.3 0.01± 15.1 0.01± 15.2 0.07±
4 0.16 7.9 0.01± 8.3 0.01± 8.1 0.1±

*Uncertainty for a is assumed to be zero because the value was provided on the slits.
*Uncertainties for x1 and x2 for Tables 1 and 2 were estimated based on the error from
measuring the distance between the minima.

Table 2:  The table below shows the distances measured between the first four minima sets, x
and the average of the two recordings.
Data number x1 0.01(mm)± x2 0.01 (mm)± x average (mm)
1 2.3     2.6   2.45 0.1±
2 7.9     7.8     7.85 0.04±
3 12.2     12.6     12.4 0.1±
4 16.8   16.9   16.85 0.04±

Sample Calculation for average distance between minima, x (used for tables 1 and 2)
xave=(x1+x2)/2

xave=(62.0+59.9)/2
xave=60.95mm

Calculation of standard deviation for average x (used for tables 1 and 2)
/2((x ) x ) )σxave = 1 1 − xave
2
+ ( 2 − xave
2 1/2

=½((62.060.95)2
+(59.960. 95)2
)1/2
σxave

=0.74246mmσxave

Analysis:
Calculation for sin ᵰ
sin ᵰ = 0.5x/(L2
+0.25x2
)½
where x is the distance measured between the minima and L is the
distance from the slit to the screen.
sin ᵰ = 0.5(60.95mm)(1000mm2
+0.25(60.95mm)2
)½

sin ᵰ = 0.030461

Propagation of uncertainty for sin ᵰ
sinθ {d/dt(sin(arctan(xave/1000)))} ]Δ = [ 2 1/2

sinθ {1000cos(arctan(xave/1000))/(xave 000000)} ]Δ = [ 2 + 1 2 1/2

sinθ {1000cos(arctan(60.95/1000))/(60.95 000000)} ]Δ = [ 2
+ 1 2 1/2

0.0001sinθΔ =

Table 3: The table below shows the calculated values of sin ᵰ  and sin ᵰ.Δ

Data number a (mm) 1/a (1/mm) sin ᵰ   sin ᵰΔ
1 0.02 50 0.030461     0.0001±
2 0.04 25 0.015948 0.0001±
3 0.08 12.5 0.0076 0.0001±
4 0.16 6.25 0.00405 0.0001±

*Uncertainty for a and 1/a are assumed to be zero because the values were provided on the slits.

Figure 1: The figure shows the graph of sin ᵰ vs. 1/a.

Figure 2: The figure below shows the linear regression analysis from the excel output of the data
from Figure 1.
   Coefficients Standard Error
Intercept 0.000296129 0.000356282
X Variable 1 0.000606657 1.23661E05

To calculate the wavelength, take the slope of Figure 1.

Calculated wavelength: 607 12 nm±
Percent error:
% error = ( exp known) / ( known ) *100λ λ λ λ
(600670)/(670)*100= 10.4%

Calculations below are for Table 4
Calculation for where d is given on the double slit and is constant for all values of m.λ
sinθL /(m .5)λ = d 2
+ 0
0.25)(0.001225)(1000 )/(0 .5)λ = ( 2
+ 0
612.50 nmλ =

Calculation propagation of uncertainty  Δ λ
λ 000(0.25 000 os(arctan(xave/1000)) sinθ/(xave 000)(m .5)))Δ = 1 * 1 * c * Δ 2 + 1 + 0
λ 000(0.25 000 os(arctan(2.45/1000)) 0.0001)/(2.45 000)(0 .5)))Δ = 1 * 1 * c * ( 2
+ 1 + 0
0.5 nmλΔ =
Calculations for sin( and sin( are the same as the previous experiment.)θ Δ )θ

Table 4 The table below shows sin( ) compared to wavelength calculated from the orderθ
integer.
Order
integer, M
sin( )θ sin(Δ )θ wavelength, λ
(nm)
(nm)  Δ λ
0 0.001225 0.0001± 612.50 0.5±
1 0.003925 0.0001± 654.16 0.2±
2 0.0062 0.0001± 619.99 0.1±
3 0.008425 0.0001± 601.76 0.07±

Average wavelength:
(612.5+654.16+619.99+601.76)/4 = 622.10 nm

Standard deviation of wavelength:
/2((λ ) λ ) )σλave = 1 1 − λave
2
+ ( 2 − λave
2 1/2

=½((612.5622.10)2
+(654.16622.10)2
+(619.99622.10)2
+(601.76 622.10)2
)1/2
σλave

= 19.609 nmσλave

Calculated wavelength: 622 20 nm±
Percent error
% error = ( exp known) / ( known ) *100λ λ λ λ
(620670)/(670)*100= 7.46%

Discussion:
Results
actual 70 0 nmλ = 6 ± 1
exp 1   00 0 nmλ = 6 ± 1
% error = 10.4 %
range = 590 610 mn mn
exp 2   20 0 nmλ = 6 ± 2
% error = 7.46 %
range= 600 640 mn mn

Conclusion:
For experiment 1, using linear regression, the experimental value of wavelength for the
laser was found to be under the actual, not even falling in the range. Most of the error occurred
while measuring the x distance between minima. Between reading mm on a ruler and measuring
the distance of light propagated most of the error. The smaller slit distance, a, was harder to
measure creating a larger estimated error.
For experiment 2, the wavelength was found using averages, but again the experimental
wavelength range was under the actual wavelength of the laser. For similar reasons as
experiment 1, the distance of x was difficult to measure leading to human error.

Post lab Questions:

1. Single Slit: Using that ƛ = 650 nm and single slit width a=0.04 mm, plot the intensity (relative
to the max Im) from α = 0 to 4π. Find the maximum value of the intensity for α between π and 2π.
Givens:
  ƛ = 650 nm,  a=0.04 mm, α = 0 to 4π
Solution:
Using the equation , intensity vs alpha is plotted below from values ranging m((sin α) /α))I = I 2

from 0 to 4π.

According to the graph, the maximum value for the intensity between π and 2π is about 0.0472 Im
at 1.43π.
2. Double Slit: Using the ƛ= 650 nm, single slit width of a=0.04 mm and the double slit distance
d= 0.25 mm, calculate the intensity of the first through tenth maxima.
Givens:

ƛ= 650 nm, a=0.04 mm, d= 0.25 mm
Solution:
sinθ mλ d =
First, solve for theta.
rcsin(λm/d)θ = a
rcsin((650 0 m)/(.25 0 ))θ = a × 1 −9
× 1 −3

Now plug the above expression into the equation.β
d sinθ/λ β = π
(.25 0 )(650 0 )((m 650 0 ))/(.25 0 ))β = π × 1 −3
× 1 −9
× ( × 1 −9
× 1 −3

The .25*103
and 650*109
terms cancel which leaves
mβ = π
Now solve for alpha using the known numbers given in the problem in the following equation.
a sinθ /λ  α = π
(.04 0 )(650 0 )m/(.25 0 )(650 0 )α = π × 1 −3
× 1 −9
× 1 −3
× 1 −9

(.04 0 )m/(.25 0 ) 16πmα = π × 1 −3
× 1 −3
= .
Finally, plug the derived expressions for alpha and beta into the intensity equation.
m(cosβ) (sinα/α)I = I 2 2

m(cos (πm))(sin(.16πm)/.16πm)I = I 2 2

The cos2
( ) term goes to 1 for all values of m, which leaves.mπ
m(sin(.16πm)/(.16πm))I = I 2

Table 5: The table below shows the calculated I values for the first ten maxima.

m I in terms of Im
1 0.918646
2 0.705638
3 0.43846
4 0.202921
5 0.054944
6 0.001771
7 0.010859
8 0.036628
9 0.047153
10 0.035895

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