Young's Double-Slit Problem

  • #1

Homework Statement


In a young's double slit experiment a thin sheet of mica is placed over one of the two slits. As a result, the center of the fringe pattern shifts by 30 bright fringes. The wavelength of light in the experiment is 480 nm and the index of refraction of the mica is 1.60. The thickness of the mica is?

Homework Equations


##d\sin\theta=m\lambda\\
\lambda_{n}=\frac{\lambda}{n}##

The Attempt at a Solution


Without considering the effect of the mica we have that the phase difference between the slits will be ##30\lambda## however I am stuck trying to figure out how to account for the additional phase difference when the mica is placed.
 

Answers and Replies

  • #2
blue_leaf77
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Remember how one obtains ##d \sin{\theta} = m\lambda##. One initially writes down the general formula (with certain approximation) for the phase difference between the rays emanating from each slit at the distant screen. How would you modify this expression of the phase difference if one source (slit) is retarded due to the presence of a thin material? Hint: in the absence of the thin material, the phase difference between the two rays from both sources at distant screen is ##\frac{2\pi}{\lambda}d \sin{\theta}##.
 
  • #3
vela
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Homework Statement


In a young's double slit experiment a thin sheet of mica is placed over one of the two slits. As a result, the center of the fringe pattern shifts by 30 bright fringes. The wavelength of light in the experiment is 480 nm and the index of refraction of the mica is 1.60. The thickness of the mica is?

Homework Equations


##d\sin\theta=m\lambda\\
\lambda_{n}=\frac{\lambda}{n}##

The Attempt at a Solution


Without considering the effect of the mica we have that the phase difference between the slits will be ##30\lambda## however I am stuck trying to figure out how to account for the additional phase difference when the mica is placed.
Consider a thickness t. How many wavelengths would fit into this thickness if the region was filled with air? How many would fit into this thickness if the region was filled with mica?
 
  • #4
Thanks for the help guys, finally managed to solve it. $$PD=30\lambda \\
PD=(\frac{t}{\lambda n}-\frac{t}{\lambda})\cdot\lambda \\
30\lambda=t_{n}-t \\
t=\frac{30\lambda}{n-1} $$
This ended up giving the correct answer.
 

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