# Diffraction Problem: Calculating Angular Width w/ Steve's Eyelashes

• tigerguy
In summary: Thanks!This changes the problem to a diffraction grating problem. The interference is from multiple slits with regular spacing between them. The diffraction from each individual slit is responsible for spreading the light in different directions, but the interference pattern is due to the path length differences from the different slits. The equations are similar, but not quite the same.Thanks again;Diffraction and interference occur for wave phenomena such as light and water waves as well as sound waves. During a lecture on diffraction, Steve's eyelids droop,meaning that he closes his eyes until he is looking out through his eyelashes. Then, a green light (lambda
tigerguy

## Homework Statement

Diffraction and interference occur for wave phenomena such as light and water waves as well as sound waves. During a lecture on diffraction, Steve's eyelids droop,meaning that he closes his eyes until he is looking out through his eyelashes. Then, a green light (lambda= 6 e -7) ini the lecture hall appears to have multiple images, and each image is blurred (wider) than when his eyes are more wide open.

What is the full angular widht (delta theta) of each of the multiple images of the light as viewed through the gap between a pair of neighboring eyelashes? Suppose that the diameter of Steve's eyelashes is 3 x 10^-6 m and the spacing between them is 6 x 10^-6 m.

## Homework Equations

I think we have to using sin(theta) = lambda/W. For the eyelashes, I think we have to use sin (theta) = 1.22 lambda/D.

## The Attempt at a Solution

So basically, I used that same formula twice, and added the theta's up to get the angle. I don't think that this approach is right, so any advice/guidance would be appreciated.

tigerguy said:

## Homework Statement

Diffraction and interference occur for wave phenomena such as light and water waves as well as sound waves. During a lecture on diffraction, Steve's eyelids droop,meaning that he closes his eyes until he is looking out through his eyelashes. Then, a green light (lambda= 6 e -7) ini the lecture hall appears to have multiple images, and each image is blurred (wider) than when his eyes are more wide open.

What is the full angular widht (delta theta) of each of the multiple images of the light as viewed through the gap between a pair of neighboring eyelashes? Suppose that the diameter of Steve's eyelashes is 3 x 10^-6 m and the spacing between them is 6 x 10^-6 m.

## Homework Equations

I think we have to using sin(theta) = lambda/W. For the eyelashes, I think we have to use sin (theta) = 1.22 lambda/D.

## The Attempt at a Solution

So basically, I used that same formula twice, and added the theta's up to get the angle. I don't think that this approach is right, so any advice/guidance would be appreciated.

The wording of the problem sounds to me like you are supposed to be analyzing the pattern of one single slit.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html#c1

What would the slit width a and the variable m be, then? I don't think we learned that in class yet. Thanks for your help!

tigerguy said:
What would the slit width a and the variable m be, then? I don't think we learned that in class yet. Thanks for your help!

The slit width is to be calcualted from the diameter and spacing of the eyelashes. m will be different for each image.

Theres a second part of the question that I'm confused about:

Steve also sees multiple images of light due to constructive interference from the light that diffracts through different gaps between his eyelashes. The most prominent secondary images occurs at an angle such that the path lengths of light from the gaps between neighboring eyelashes to the retina differ by one wavelength so that their interference is constructive. What is the angular separation between the primary image (seen with wide-open eyes) and the nearest secondary image (seen through the eyelashes)?

What is the best way to approach this problem? Are we just finding the angles of both, and just subtracting to find the difference? I'm pretty confused, so I'd appreciate any help.

tigerguy said:
Theres a second part of the question that I'm confused about:

Steve also sees multiple images of light due to constructive interference from the light that diffracts through different gaps between his eyelashes. The most prominent secondary images occurs at an angle such that the path lengths of light from the gaps between neighboring eyelashes to the retina differ by one wavelength so that their interference is constructive. What is the angular separation between the primary image (seen with wide-open eyes) and the nearest secondary image (seen through the eyelashes)?

What is the best way to approach this problem? Are we just finding the angles of both, and just subtracting to find the difference? I'm pretty confused, so I'd appreciate any help.

This changes the problem to a diffraction grating problem. The interference is from multiple slits with regular spacing between them. The diffraction from each individual slit is responsible for spreading the light in different directions, but the interference pattern is due to the path length differences from the different slits. The equations are similar, but not quite the same.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/grating.html

Thanks again;

I found 2 angles for the last problem - would the difference be the angular separation between the primary and secondary images?

Thanks

tigerguy said:
Thanks again;

I found 2 angles for the last problem - would the difference be the angular separation between the primary and secondary images?

Thanks

That's the way it sounds to me. The primary image formed when looking through the eyelash "grating" is in the same position as the one image formed with no grating.

## 1. What is the diffraction problem?

The diffraction problem refers to the phenomenon of light bending or spreading out when it encounters an obstacle or passes through a narrow opening. This can cause interference patterns and affect the quality of images produced by optical devices.

## 2. How is the angular width calculated?

The angular width in the diffraction problem can be calculated using the formula w = λ/D, where w is the angular width, λ is the wavelength of light, and D is the size of the obstacle or opening. This formula is known as the Rayleigh criterion.

## 3. Who is Steve and why are his eyelashes involved in this problem?

Steve is a fictional character often used in examples to explain the concept of diffraction. In this particular problem, we are using the width of Steve's eyelashes as the size of the obstacle to calculate the angular width.

## 4. What is the significance of calculating the angular width in the diffraction problem?

Calculating the angular width helps us understand the diffraction effect on light passing through small openings or obstacles. It also helps in determining the resolution of optical devices, such as microscopes and telescopes.

## 5. Can the diffraction problem be solved for any type of wave?

Yes, the diffraction problem can be solved for any type of wave, including light, sound, and water waves. The formula for calculating the angular width may vary slightly depending on the type of wave, but the concept remains the same.

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