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iris [78.8K]
3 years ago
6

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

central maximum. at what angles do the second bright spots occur?
Physics
1 answer:
o-na [289]3 years ago
3 0
The equation we use is mλ=dsinθ for intensity maximas. We are given at the first maximum (m=1), it occurs at 17.8 degrees. Thus we can solve for d by substituting known values into our equation.

(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m

Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.

(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ

θ=37.69 degrees

Hopes this helps!

P.S. I hope this is right. If not sorry in advance.
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Kobotan [32]

Answer:

Se the explanation below

Explanation:

We do not feel these forces of these bodies, because they are very small compared to the force of Earth's attraction. Although its mass is greater than that of a human being, its mass is not compared to the Earth's mass. In order to understand this problem we will use numerical data and the universal gravitation formula, to give validity to the explanation.

<u>Force exerted by the Earth on a human being</u>

<u />

F=G*\frac{m_{1}*m_{2}}{r^2}

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 5.97*10^24[kg]

r = distance from the center of the earth to the surface or earth radius = 6371 *10^3 [m]

<u />

Now replacing we have

F = 6.673*10^{-11} *\frac{80*5.97*10^{24}}{(6371*10^{3})^{2}  } \\F = 785[N]

<u>Force exerted by a building on a human being</u>

<u />

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 300000 [ton] = 300 *10^6[kg]

r = distance from the building to the person = 2[m]

F = 6.673*10^{-11}*\frac{80*300*10^6}{2^{2} }  \\F= 0.4 [N]

As we can see the force exerted by the Earth is 2000 times greater than that exerted by a building with the proposed data.

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2 years ago
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3 years ago
Read 2 more answers
What is the speed of sound at 33 °C (m/s)? For a frequency of 5 kHz, how large do you expect the wavelength to be (m)?
Vedmedyk [2.9K]

Answer:

-The speed of sound at 33°C is 362.8 m/s.

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Explanation:

let v = 343 m/s be the speed of sound.

let T be the temperature.

then the speed of sound V, at 33°C is given by:

V = v + 0.6×T

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Therefore, the speed of sound at 33°C is 362.8 m/s.

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λ = V/f

  = (362.8)/(5000)

  = 0.07256 m

Therefore, the wavelength at a frequency at 5 kHz is 0.07256 m .

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3 years ago
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