Classroom Demo of Coherent Light Beam on Slits: Find Fringe Distance

In summary, a beam of coherent light of wavelength 550 nm is incident perpendicularly onto a pair of slits with a width of 1.2 x 10^-6 m and a distance of 1.8x10^-5 m between the centers. The class observes light and dark fringes on a screen 2.2 m away. To calculate the frequency of the light, the equation f = c/\lambda is used, resulting in a frequency of 5.45*10^14 Hz. To find the distance between two adjacent dark fringes, the equation ydark = [λL/d][m+.5] is used, where m is the order of the fringe. The frequency of the light in
  • #1
science.girl
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0

Homework Statement


In a classroom demonstration, a beam of coherent light of wavelength 550 nm is incident perpendicularly onto a pair of slits. Each slit has a width w of 1.2 x 10^-6 m, and the distance d between the centers of the slits is 1.8x10^-5 m. The class observes light and dark fringes on a screen that is a distance L of 2.2 m from the slits. Your notebook shows the following setup for the demonstration.

See page 11: http://apcentral.collegeboard.com/apc/public/repository/ap09_frq_physics_b.pdf

(a) Calculate the frequency of the light.

(b) Calculate the distance between two adjacent dark fringes on the screen.

The entire apparatus is now immersed in a transparent fluid having index of refraction 1.4.
(c) What is the frequency of the light in the transparent fluid?

(d) Does the distance between the dark fringes increase, decrease, or remain the same?

_____ Increase _____ Decrease _____ Remain the same

Explain your reasoning.

Homework Equations


f = c/[tex]\lambda[/tex]

ydark = [[tex]\lambda[/tex]L/d][m+.5]

The Attempt at a Solution



A) f = c/[tex]\lambda[/tex]
f = (3.0*10^8 m/s)/(5.5*10^(-7) m) = 165Hz

B) ydark = [[tex]\lambda[/tex]L/d][m+.5]
Hmm... would you have to find a [tex]\Delta[/tex]y, though? How would you do this?
 
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  • #2
Find ym and ym+1 for dark fringes and find the difference.
What is the relation between the refractive index of the material and wavelength of light traveling through?
 
  • #3
science.girl said:

Homework Equations


f = c/[tex]\lambda[/tex]

ydark = [[tex]\lambda[/tex]L/d][m+.5]

The Attempt at a Solution



A) f = c/[tex]\lambda[/tex]
f = (3.0*10^8 m/s)/(5.5*10^(-7) m) = 165Hz
Careful with handling exponents: 8 - (-7) = ? You can tell that this number is way off.

B) ydark = [[tex]\lambda[/tex]L/d][m+.5]
Hmm... would you have to find a [tex]\Delta[/tex]y, though? How would you do this?
How would you find the y position of successive dark lines? (Hint: what is m?) How would you find the difference in the y position of these dark lines?

AM
 
  • #4
Andrew Mason said:
Careful with handling exponents: 8 - (-7) = ? You can tell that this number is way off.

How would you find the y position of successive dark lines? (Hint: what is m?) How would you find the difference in the y position of these dark lines?

AM

Oops! (A) should be 5.45*10^14 Hz.

For (B):
rl.bhat says to "Find ym and ym+1 for dark fringes and find the difference."
Hmm... I'm going to think about this one some more. Any hints, maybe?
I'll keep you updated.
 
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  • #5
science.girl said:
For (B):
rl.bhat says to "Find ym and ym+1 for dark fringes and find the difference."
Hmm... I'm going to think about this one some more. Any hints, maybe?

Use the equation you wrote earlier:
science.girl said:
ydark = [λL/d][m+.5]

Evaluate the RHS for m and also for m+1. Subtract the two expressions and see what you get.
 
  • #6
Redbelly98 said:
Use the equation you wrote earlier:


Evaluate the RHS for m and also for m+1. Subtract the two expressions and see what you get.

Oh! I understand now! Thank you very much.

For (c), I used the v = c/n, and used the velocity in the equation v = f[tex]\lambda[/tex] to determine the frequency. Is this correct?

For (d), under what conditions would the dark fringes move closer or farther away? What should I be analyzing?
 
  • #7
(c) Correct

(d) Two things to think about here:

1. What happens to the wavelength when the medium is changed from n=1 to n=1.4?
2. How does the fringe spacing change when wavelength is either increased or decreased?
 
  • #8
Perfect! Thank you.
 

1. What is a coherent light beam?

A coherent light beam is a type of light that has consistent and predictable properties. It is produced by a single light source, such as a laser, and has a consistent wavelength, direction, and phase.

2. How does a coherent light beam create fringes on slits?

When a coherent light beam passes through a narrow slit, it diffracts and spreads out, creating a pattern of dark and light fringes. This is due to the interference of the light waves as they pass through the slit.

3. What is the fringe distance in a coherent light beam on slits demo?

The fringe distance refers to the distance between adjacent dark or light fringes on the screen where the coherent light beam is projected. It is a measure of the interference pattern created by the diffraction of the light through the slits.

4. What factors can affect the fringe distance in a coherent light beam on slits?

The fringe distance can be affected by several factors, including the wavelength of the light, the distance between the slits, the distance between the slits and the screen, and the size of the slits. Any changes in these factors can alter the interference pattern and therefore, the fringe distance.

5. What are the practical applications of the coherent light beam on slits experiment?

The coherent light beam on slits experiment has several practical applications in science and technology. It can be used to study the properties of light and its behavior, as well as to calibrate and test optical instruments such as telescopes and microscopes. It is also used in various technologies, such as barcode scanners, holography, and optical communications.

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