How Do Interference Patterns and Diffraction Determine Angles in Physics?

In summary, the conversation discusses two problems. The first problem involves a sodium-vapor street lamp and the formation of an interference pattern on a distant wall. The second problem involves using a diffraction-grating spectroscope to determine the chemical composition of a star. The angle of the fourth-order minimum is being asked for in problem 6, while the angle at which the second-order spectral line for wavelength 1 occurs is being asked for in problem 15. The equations used in both problems involve wavelength, slit separation, and angle.
  • #1
mustang
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Problem 6. A sodium-vapor street lamp produces light that is nearly monochromatic. If the light shines on a wooden door in which there are two straight, parallel cracks, an interfernece pattern will form on a distant wall behind the door. The slits have a separation of 0.3740mm, and the second-order maximum occurs at an angle of 0.18046 degrees from the central maximum.
determine the angle of the fourth-order minimum. Answer in degrees.
Would it be:
d sinθ = (m+1/2)λ
(0.3740*10^-3)sinθ = (m+1/2)λ does m=3?
Problem 15. By attaching a diffraction-grating spectroscope to an astronomical telescope, one can measure the spectral lines from a start and determine the start's chemical composition. Assume the grating has 3224 slits/cm. The wavelengths of the star's light are wavelength_1=463.200nm, wavelength_2=640.500 nm, and wavelength_3=704.700 nm.

Find the angle at which the second-order spectral line for wavelength 1 occurs. Answer in deg
rees.
 
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  • #2
Problem 6:
What are you going to use for λ?

Problem 15:
This is pretty much just plugging numbers into a equation. What's your question?
 
  • #3


To solve this problem, we can use the formula d sinθ = mλ, where d is the slit separation, θ is the angle at which the spectral line occurs, m is the order of the line, and λ is the wavelength.

For the first part of the problem, we need to find the angle of the fourth-order minimum. Since we are given the slit separation (d = 0.3740mm) and the second-order maximum angle (θ = 0.18046 degrees), we can rearrange the formula to solve for the fourth-order minimum angle (θ).

d sinθ = (m+1/2)λ
0.3740mm * sinθ = (4+1/2) * λ
sinθ = 2.5 * (λ / 0.3740mm)

Now, we plug in the value of the wavelength (λ = 463.200nm) and convert it to meters (0.463200m) to get:

sinθ = 2.5 * (0.463200m / 0.3740mm)
sinθ = 3.103 * 10^-3

We can use a calculator to find the inverse sine of this value, which gives us θ = 0.178 degrees. Therefore, the angle of the fourth-order minimum is 0.178 degrees.

For the second part of the problem, we need to find the angle at which the second-order spectral line for wavelength 1 occurs. Again, we can use the formula d sinθ = mλ, and rearrange it to solve for θ.

d sinθ = mλ
0.3740mm * sinθ = 2 * (463.200nm / 3224 slits/cm)

We can convert the slit separation to meters (0.3740mm = 3.740 * 10^-4m) and the slits/cm to slits/m (3224 slits/cm = 3.224 * 10^4 slits/m) to get:

3.740 * 10^-4m * sinθ = 2 * (0.463200m / 3.224 * 10^4 slits/m)
sinθ = 3.103 * 10^-3

Again, using a calculator, we find that θ = 0.178 degrees. Therefore, the angle at which
 

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