Double Slit Experiment - White Lite & Minimum Angular Displacement

[tnbstudent]

Homework Statement

A two-component beam of light, consisting only of two wavelengths 650 nm and 520 nm, is used to obtain interference fringes in a double-slit experiment. The separation between the centers of the slits is 2.43×10−4 m and the distance of the plane of the slits from the screen is 720 cm. What is the smallest angular displacement from the central maximum where the centers of the bright fringes due to the two colors will coincide?

Homework Equations

sinθ=(m/d)λ
m is number of fringes
d is the width of the slit
sinθ=a/D
a is the distance from the center on the other side
D is the distance the slit is to the screen

λ1=650 nm
λ2 =520 nm
D=7.2 m
d= 2.43×10−4 m

The Attempt at a Solution

I set up a series of equations like this:
(m1/d)λ1 = (m2/d)λ2 = sinθ = a/D
(m1/2.43×10−4 m)*650nm = (m2/2.43×10−4 m)*520nm = a/7.2m
a=1.926*10^-2 *(m1) = 1.54*10^-2m* (m2)
I found that when m1 = 4 and m2 = 5, a =7.704*10^-2m
sinθ=a/d
sinθ=.0107
θ= .61°

Thanks for any help that will point me in the right direction

 

[anathema]

!



Your approach is correct and your final answer of 0.61° is correct. However, I would like to provide some additional clarification and explanation for your solution.

Firstly, your initial setup of the equations is correct. The only thing I would like to point out is that the value for D should be 7.2 m instead of 720 cm. This is because all the other values given in the problem are in meters and it is important to be consistent with units when solving equations.

Next, the values for m1 and m2 represent the number of fringes for each wavelength, not the number of fringes for both wavelengths combined. Therefore, the equation should be (m1/d)λ1 = (m2/d)λ2 = sinθ = a/D. This means that the number of fringes for each wavelength will be different and will contribute to the overall displacement of the bright fringes.

When solving for a, you can set the equations (m1/d)λ1 = (m2/d)λ2 and a/D equal to each other to find the relation between m1 and m2. In this case, you will get m1 = (5/4)m2. This means that for every 5 fringes of 520 nm, there will be 4 fringes of 650 nm, resulting in a total displacement of 9 fringes. This is why your final answer of 0.61° is correct, as it represents the angular displacement for 9 fringes.

Overall, your solution is correct and well thought out. Just remember to be consistent with units and to consider the number of fringes for each wavelength separately. Keep up the good work!