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-BARSIC- [3]
2 years ago
11

A lab technician uses laser light with a wavelength of 670 nm to test a diffraction grating. When the grating is 40.0 cm from th

e screen, the first-order maxima appear 6.00 cm from the center of the pattern. How many lines per millimeter does this grating have?
Physics
1 answer:
Juliette [100K]2 years ago
8 0

Answer:

N = 221.4 lines / mm

Explanation:

Given:

- The wavelength of the source λ = 670 nm

- Distance of the grating from screen B = 40.0 cm

- The distance of first bright fringe from central order P = 6.0 cm

Find:

How many lines per millimeter does this grating have?

Solution:

- The derived results from Young's experiment that relates the order of bright fringes about the central order is given by:

                                          sin (Q) = n*λ*N

Where,

n is the order number 0, 1 , 2, 3 , ....

λ  is the wavelength of the light source

Q is the angle of sweep respective fringe from central order

N is the number of lines/mm the grating has

- We will first compute the length along which the light travels for the first bright fringe:

                                            L^2 = P^2 + B^2

                                            L^2 = 40^2 + 6^2

                                            L^2 = 1636

                                            L = 40.45 cm  

- Now calculate the sin(Q) that the fringe makes with the central order:

                                            sin (Q) = P / L

                                            sin (Q) = 6 / 40.45

- Now we will use the derived results:

                                           N = sin(Q) / n*λ

  Where, n = 1 - First order

  Plug values in                N = (6 / 40.45) / (670 *10^-6)

                                          N = 221.4 lines / mm

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Answer:

H = 109.14 cm

Explanation:

given,                                                            

Assume ,                                                            

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earnstyle [38]

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Explanation:

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So, the distance of the listener from speaker B is ...

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_____

The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

_____

<em>Comment on the speed of sound</em>

We don't know what speed you are to use for the speed of sound. We have used 343 m/s. Some sources use 340 m/s, which will give a result different by 2 or 3 cm.

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