Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 25, Problem 43P

(a)

To determine

The number of wavelengths present in the spectrum of the light source.

(a)

Expert Solution
Check Mark

Answer to Problem 43P

There are two spectral lines with wavelengths 449.2nm and 665.1nm.

Explanation of Solution

Write the expression to calculate the slit width.

  d=1n

Here, d is the slit width and n is the number of slits.

Substitute 5000.0cm1 for n in the above equation to calculate d.

  d=15000.0cm1=2.00×104cm(102m1cm)=2.00×106m

Write the expression to calculate the first order wavelength for the smallest angle.

  λ=dsinθm

Here, λ is the wavelength.

Substitute 2.00×106m for d, 12.98° for θ and 1 for m in the above equation to calculate λ.

  λ=(2.00×106m)sin12.98°1=0.4492×106m=449.2×109m(1nm109m)=449.2nm

Write the expression for the angle of second order lines for the above wavelength.

  θ=sin1(2λd)

Substitute 449.2nm for λ and 2.00×106m for d in the above equation to calculate θ.

  θ=sin1(2(449.2nm(109m1nm))2.00×106m)=26.69°

Write the expression for the angle of third order lines for the above wavelength.

  θ=sin1(3λd)

Substitute 449.2nm for λ and 2.00×106m for d in the above equation to calculate θ.

  θ=sin1(3(449.2nm(109m1nm))2.00×106m)=42.36°

Write the expression for the angle of fourth order lines for the above wavelength.

  θ=sin1(4λd)

Substitute 449.2nm for λ and 2.00×106m for d in the above equation to calculate θ.

  θ=sin1(4(449.2nm(109m1nm))2.00×106m)=63.95°

Thus, these angles belong to the second order wavelengths and these angles could be omitted.

Write the expression to calculate the first order wavelength for the angle 19.0°.

  λ=dsinθm

Substitute 2.00×106m for d, 19.0° for θ and 1 for m in the above equation to calculate λ.

  λ=(2.00×106m)sin19.0°1=0.6510×106m=665.1×109m(1nm109m)=665.1nm

Write the expression for the angle of second order lines for the above wavelength.

  θ=sin1(2λd)

Substitute 665.1nm for λ and 2.00×106m for d in the above equation to calculate θ.

  θ=sin1(2(665.1nm(109m1nm))2.00×106m)=40.6°

Write the expression for the angle of third order lines for the above wavelength.

  θ=sin1(3λd)

Substitute 665.1nm for λ and 2.00×106m for d in the above equation to calculate θ.

  θ=sin1(3(665.1nm(109m1nm))2.00×106m)=77.6°

Thus, the angle corresponding the wavelength could be omitted.

Conclusion:

Therefore, there are two spectral lines with wavelengths 449.2nm and 665.1nm.

(b)

To determine

The number of spectral lines on one side of the central maximum.

(b)

Expert Solution
Check Mark

Answer to Problem 43P

The number of spectral lines is 19.

Explanation of Solution

Write the expression to calculate the slit width.

  d=1n

Here, d is the slit width and n is the number of slits.

Substitute 2000.0cm1 for n in the above equation to calculate d.

  d=12000.0cm1=5.00×104cm(102m1cm)=5.00×106m

Write the expression to calculate order corresponding to the wavelength 449.2nm.

  m=dsinθλ

Substitute 90.0° for θ, 5.00×106m for d and 449.2nm for λ in the above equation to calculate m.

  m=(5.00×106m)sin90.0°449.2nm(109m1nm)=11.111

Write the expression to calculate order corresponding to the wavelength 665.1nm.

  m=dsinθλ

Substitute 90.0° for θ, 5.00×106m for d and 665.1nm for λ in the above equation to calculate m.

  m=(5.00×106m)sin90.0°665.1nm(109m1nm)=7.58

Thus, the total number of spectral lines is 11+8=19.

Conclusion:

Therefore, the number of spectral lines is 19.

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