Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 27, Problem 60P

(a)

To determine

The angle of the second order ray in diffraction grating.

(a)

Expert Solution
Check Mark

Answer to Problem 60P

The angle of the second order ray in the diffraction grating that has 400 grooves/mm is 25.6° .

Explanation of Solution

Given info: The number of grooves per mm in the diffraction grating is 400, the incident wavelength is 541nm

Formula to calculate angle of mth order ray is ,

θ=sin1(mλd) (1)

Here,

d is the grid spacing.

θ angle of the mth order ray.

λ wavelength of the incident light.

First calculate the grid spacing, formula to calculate grid spacing is,

d=1numberofgroovespermetre (2)

Substitute 400mm for numberofgroovespermetre in equation (2)

d=1numberofgroovespermetre=1400mm=1400mm×1m103mm=2.50×106m (3)

Substitute 2.50×106m for d , 2 for m and 541nm for λ (1),

θ=sin1(mλd)=sin1(2×541nm2.50×106m)=sin1(2×541nm×1m109nm2.50×106m)=25.6°

Conclusion:

Therefore, The angle of the second order ray in the diffraction grating that has 400 grooves/mm is 25.6° .

(b)

To determine

The change in the angle of the second order ray in the diffraction grating if the apparatus is in water.

(b)

Expert Solution
Check Mark

Answer to Problem 60P

If the apparatus is submerged in the water the angle of the second order ray in the diffraction grating that has 400grooves/mm is 18.9°

Explanation of Solution

Given info: The number of grooves per mm in the diffraction grating is 400, the incident wavelength is 541nm The refractive index of the water is 1.33

From equation (1) formula to calculate angle of mth order ray in water is

θ=sin1(mλd) (4)

Here,

λ is the wavelength of the light in water

The whole apparatus is now submerged into water therefore the wavelength of the incident light will be reduced to factor of the refractive index of water.

The new incident wavelength is

λ=λn (5)

Here,

λ is the wavelength of the incident light.

n is the refractive index of water.

Substitute 1.33 for n , 541nm for λ in equation (5),

λ=λn=541nm×1m109nm1.33=4.06×107m (6)

From equation (6), substitute 4.06×107m for λ 2.50×106m for d , 2 for m , in equation (4).

θ=sin1(mλd)=sin1(2×4.06×107m2.50×106m)=18.9°

Conclusion:

Therefore, if the apparatus is submerged in the water the angle of the second order ray in the diffraction grating that has 400 grooves/mm is 18.9°

(c)

To determine

To show: The two diffracted rays from part (a) and part (b) are related through the law of refraction.

(c)

Expert Solution
Check Mark

Answer to Problem 60P

The relation between the diffracted rays for the part (a) and part b is n2dsinθ=n1dsinθ satisfying the law of refraction.

Explanation of Solution

 for the part (a) from equation (1),

θ=sin1(mλd)dsinθ=mλ (7)

For second order ray substitute 2 for m in equation (7)

dsinθ=mλdsinθ=2λ (8)

From equation (4) for the second part when the apparatus is submerged in water,

dsinθ=mλ (9)

Form equation (5) Substitute λn for λ in equation (9)

dsinθ=mλdsinθ=mλnndsinθ=mλ

For second order, ray substitute 2 for m in the above equation.

ndsinθ=2λ (10)

Compare equation (8) and (10).

ndsinθ=dsinθ

Substitute n2 for n in left hand side of the equation n1 for 1 in the right hand side of the above equation.

ndsinθ=dsinθn2dsinθ=1×dsinθn2dsinθ=n1dsinθ (11)

Here,

n2 is the refractive index of the material

n1 is the refractive index of the air.

From equation (11), it is evident that it can be expressed in terms of law of refraction.

Conclusion:

Therefore, the two diffracted rays of part (a) and part (b) is related through law of refraction.

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Chapter 27 Solutions

Principles of Physics: A Calculus-Based Text

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