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Mashcka [7]
3 years ago
8

If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th

e path difference?
nm




How many orders of bright lines does this equal for red light with a wavelength of 650 nm?


wavelengths
Physics
2 answers:
Trava [24]3 years ago
8 0

Answer:

-  path differnce = 2.18*10^-6

-  1538 lines

Explanation:

- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:

path\ difference\ = dsin\theta           (1)

d: separation between slits = 0.50mm = 0.50*10^-3 m

θ: angle of a diffraction = 0.25°

Then, the path difference is:

path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m

- The maximum number of bright lines are calculated by using the following formula:

m\lambda = dsin\theta           (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769

The maxium number of bright lines are twice the previous result, that is, 1538 lines

Masja [62]3 years ago
5 0

Answer:

a. 2200

b. 3

Explanation:

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6 0
3 years ago
Read 2 more answers
Add to
TEA [102]

Answer:

1) x_total = 2.7 km=2700 m, 2)   t_total = 1926 s,  3)   v_avg = 1.40 m / s

Explanation:

1) To solve this uniform velocity problem, we must find the displacement of each part and add them.

        x_total = x₁ + x₂

        x_total = 0.7 + 2.0

        x_total = 2.7 km

2) how long does it take for this tour

         

truck

        v = x₁ / t₁

        t₁ = x₁ / v

        t₁ = 0.7 / 20

        t₁ = 0.035 h

Let's reduce the time to the SI system

         t₁ = 0.035 h (3600 s / 1h) = 126 s

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         t₂ = 30 min (60 s / 1min) = 1800 s

the total time is

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        t_total = 126 + 1800

        t_total = 1926 s

3) the average velocity is defined as the displacement in the inerval between time

        v_avg = x_total / t_total

         v_avg = 2700/1926

         v_avg = 1.40 m / s

8 0
3 years ago
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