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

Question

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Answer 1

Question

Chapter 25, Problem 43P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 43P

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

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 5000.0 cm−1 for n in the above equation to calculate d.

d=15000.0 cm−1=2.00×10−4 cm(10−2 m1 cm)=2.00×10−6 m

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

λ=dsinθm

Here, λ is the wavelength.

Substitute 2.00×10−6 m for d, 12.98° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin12.98°1=0.4492×10−6 m=449.2×10−9 m(1 nm10−9 m)=449.2 nm

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

θ=sin−1(2λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(449.2 nm(10−9 m1 nm))2.00×10−6 m)=26.69°

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

θ=sin−1(3λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(449.2 nm(10−9 m1 nm))2.00×10−6 m)=42.36°

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

θ=sin−1(4λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(4(449.2 nm(10−9 m1 nm))2.00×10−6 m)=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 2.00×10−6 m for d, 19.0° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin19.0°1=0.6510×10−6 m=665.1×10−9 m(1 nm10−9 m)=665.1 nm

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

θ=sin−1(2λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(665.1 nm(10−9 m1 nm))2.00×10−6 m)=40.6°

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

θ=sin−1(3λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(665.1 nm(10−9 m1 nm))2.00×10−6 m)=77.6°

Thus, the angle corresponding the wavelength could be omitted.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

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 2000.0 cm−1 for n in the above equation to calculate d.

d=12000.0 cm−1=5.00×10−4 cm(10−2 m1 cm)=5.00×10−6 m

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 449.2 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°449.2 nm(10−9 m1 nm)=11.1≈11

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 665.1 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°665.1 nm(10−9 m1 nm)=7.5≈8

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

Conclusion:

Therefore, the number of spectral lines is 19.

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Answer 2

Question

Chapter 25, Problem 43P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 43P

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

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 5000.0 cm−1 for n in the above equation to calculate d.

d=15000.0 cm−1=2.00×10−4 cm(10−2 m1 cm)=2.00×10−6 m

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

λ=dsinθm

Here, λ is the wavelength.

Substitute 2.00×10−6 m for d, 12.98° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin12.98°1=0.4492×10−6 m=449.2×10−9 m(1 nm10−9 m)=449.2 nm

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

θ=sin−1(2λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(449.2 nm(10−9 m1 nm))2.00×10−6 m)=26.69°

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

θ=sin−1(3λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(449.2 nm(10−9 m1 nm))2.00×10−6 m)=42.36°

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

θ=sin−1(4λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(4(449.2 nm(10−9 m1 nm))2.00×10−6 m)=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 2.00×10−6 m for d, 19.0° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin19.0°1=0.6510×10−6 m=665.1×10−9 m(1 nm10−9 m)=665.1 nm

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

θ=sin−1(2λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(665.1 nm(10−9 m1 nm))2.00×10−6 m)=40.6°

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

θ=sin−1(3λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(665.1 nm(10−9 m1 nm))2.00×10−6 m)=77.6°

Thus, the angle corresponding the wavelength could be omitted.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

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 2000.0 cm−1 for n in the above equation to calculate d.

d=12000.0 cm−1=5.00×10−4 cm(10−2 m1 cm)=5.00×10−6 m

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 449.2 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°449.2 nm(10−9 m1 nm)=11.1≈11

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 665.1 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°665.1 nm(10−9 m1 nm)=7.5≈8

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

Conclusion:

Therefore, the number of spectral lines is 19.

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Answer 3

Question

Chapter 25, Problem 43P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 43P

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

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 5000.0 cm−1 for n in the above equation to calculate d.

d=15000.0 cm−1=2.00×10−4 cm(10−2 m1 cm)=2.00×10−6 m

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

λ=dsinθm

Here, λ is the wavelength.

Substitute 2.00×10−6 m for d, 12.98° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin12.98°1=0.4492×10−6 m=449.2×10−9 m(1 nm10−9 m)=449.2 nm

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

θ=sin−1(2λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(449.2 nm(10−9 m1 nm))2.00×10−6 m)=26.69°

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

θ=sin−1(3λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(449.2 nm(10−9 m1 nm))2.00×10−6 m)=42.36°

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

θ=sin−1(4λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(4(449.2 nm(10−9 m1 nm))2.00×10−6 m)=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 2.00×10−6 m for d, 19.0° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin19.0°1=0.6510×10−6 m=665.1×10−9 m(1 nm10−9 m)=665.1 nm

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

θ=sin−1(2λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(665.1 nm(10−9 m1 nm))2.00×10−6 m)=40.6°

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

θ=sin−1(3λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(665.1 nm(10−9 m1 nm))2.00×10−6 m)=77.6°

Thus, the angle corresponding the wavelength could be omitted.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

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 2000.0 cm−1 for n in the above equation to calculate d.

d=12000.0 cm−1=5.00×10−4 cm(10−2 m1 cm)=5.00×10−6 m

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 449.2 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°449.2 nm(10−9 m1 nm)=11.1≈11

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 665.1 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°665.1 nm(10−9 m1 nm)=7.5≈8

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

Conclusion:

Therefore, the number of spectral lines is 19.

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Answer 4

Question

Chapter 25, Problem 43P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 43P

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

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 5000.0 cm−1 for n in the above equation to calculate d.

d=15000.0 cm−1=2.00×10−4 cm(10−2 m1 cm)=2.00×10−6 m

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

λ=dsinθm

Here, λ is the wavelength.

Substitute 2.00×10−6 m for d, 12.98° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin12.98°1=0.4492×10−6 m=449.2×10−9 m(1 nm10−9 m)=449.2 nm

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

θ=sin−1(2λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(449.2 nm(10−9 m1 nm))2.00×10−6 m)=26.69°

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

θ=sin−1(3λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(449.2 nm(10−9 m1 nm))2.00×10−6 m)=42.36°

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

θ=sin−1(4λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(4(449.2 nm(10−9 m1 nm))2.00×10−6 m)=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 2.00×10−6 m for d, 19.0° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin19.0°1=0.6510×10−6 m=665.1×10−9 m(1 nm10−9 m)=665.1 nm

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

θ=sin−1(2λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(665.1 nm(10−9 m1 nm))2.00×10−6 m)=40.6°

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

θ=sin−1(3λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(665.1 nm(10−9 m1 nm))2.00×10−6 m)=77.6°

Thus, the angle corresponding the wavelength could be omitted.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

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 2000.0 cm−1 for n in the above equation to calculate d.

d=12000.0 cm−1=5.00×10−4 cm(10−2 m1 cm)=5.00×10−6 m

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 449.2 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°449.2 nm(10−9 m1 nm)=11.1≈11

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 665.1 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°665.1 nm(10−9 m1 nm)=7.5≈8

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

Conclusion:

Therefore, the number of spectral lines is 19.

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Answer 5

Chapter 25, Problem 43P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 43P

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

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 5000.0 cm−1 for n in the above equation to calculate d.

d=15000.0 cm−1=2.00×10−4 cm(10−2 m1 cm)=2.00×10−6 m

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

λ=dsinθm

Here, λ is the wavelength.

Substitute 2.00×10−6 m for d, 12.98° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin12.98°1=0.4492×10−6 m=449.2×10−9 m(1 nm10−9 m)=449.2 nm

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

θ=sin−1(2λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(449.2 nm(10−9 m1 nm))2.00×10−6 m)=26.69°

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

θ=sin−1(3λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(449.2 nm(10−9 m1 nm))2.00×10−6 m)=42.36°

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

θ=sin−1(4λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(4(449.2 nm(10−9 m1 nm))2.00×10−6 m)=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 2.00×10−6 m for d, 19.0° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin19.0°1=0.6510×10−6 m=665.1×10−9 m(1 nm10−9 m)=665.1 nm

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

θ=sin−1(2λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(665.1 nm(10−9 m1 nm))2.00×10−6 m)=40.6°

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

θ=sin−1(3λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(665.1 nm(10−9 m1 nm))2.00×10−6 m)=77.6°

Thus, the angle corresponding the wavelength could be omitted.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

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 2000.0 cm−1 for n in the above equation to calculate d.

d=12000.0 cm−1=5.00×10−4 cm(10−2 m1 cm)=5.00×10−6 m

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 449.2 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°449.2 nm(10−9 m1 nm)=11.1≈11

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 665.1 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°665.1 nm(10−9 m1 nm)=7.5≈8

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

Conclusion:

Therefore, the number of spectral lines is 19.

Answer 6

Chapter 25, Problem 43P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 43P

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

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 5000.0 cm−1 for n in the above equation to calculate d.

d=15000.0 cm−1=2.00×10−4 cm(10−2 m1 cm)=2.00×10−6 m

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

λ=dsinθm

Here, λ is the wavelength.

Substitute 2.00×10−6 m for d, 12.98° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin12.98°1=0.4492×10−6 m=449.2×10−9 m(1 nm10−9 m)=449.2 nm

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

θ=sin−1(2λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(449.2 nm(10−9 m1 nm))2.00×10−6 m)=26.69°

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

θ=sin−1(3λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(449.2 nm(10−9 m1 nm))2.00×10−6 m)=42.36°

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

θ=sin−1(4λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(4(449.2 nm(10−9 m1 nm))2.00×10−6 m)=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 2.00×10−6 m for d, 19.0° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin19.0°1=0.6510×10−6 m=665.1×10−9 m(1 nm10−9 m)=665.1 nm

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

θ=sin−1(2λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(665.1 nm(10−9 m1 nm))2.00×10−6 m)=40.6°

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

θ=sin−1(3λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(665.1 nm(10−9 m1 nm))2.00×10−6 m)=77.6°

Thus, the angle corresponding the wavelength could be omitted.

Conclusion:

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

Answer 7

Chapter 25, Problem 43P

(a)

To determine

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

Answer 8

(a)

Expert Solution

Answer to Problem 43P

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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 5000.0 cm−1 for n in the above equation to calculate d.

d=15000.0 cm−1=2.00×10−4 cm(10−2 m1 cm)=2.00×10−6 m

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

λ=dsinθm

Here, λ is the wavelength.

Substitute 2.00×10−6 m for d, 12.98° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin12.98°1=0.4492×10−6 m=449.2×10−9 m(1 nm10−9 m)=449.2 nm

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

θ=sin−1(2λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(449.2 nm(10−9 m1 nm))2.00×10−6 m)=26.69°

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

θ=sin−1(3λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(449.2 nm(10−9 m1 nm))2.00×10−6 m)=42.36°

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

θ=sin−1(4λd)

Substitute 449.2 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(4(449.2 nm(10−9 m1 nm))2.00×10−6 m)=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 2.00×10−6 m for d, 19.0° for θ and 1 for m in the above equation to calculate λ.

λ=(2.00×10−6 m)sin19.0°1=0.6510×10−6 m=665.1×10−9 m(1 nm10−9 m)=665.1 nm

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

θ=sin−1(2λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(2(665.1 nm(10−9 m1 nm))2.00×10−6 m)=40.6°

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

θ=sin−1(3λd)

Substitute 665.1 nm for λ and 2.00×10−6 m for d in the above equation to calculate θ.

θ=sin−1(3(665.1 nm(10−9 m1 nm))2.00×10−6 m)=77.6°

Thus, the angle corresponding the wavelength could be omitted.

Conclusion:

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

Answer 13

(b)

To determine

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

(b)

Expert Solution

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 2000.0 cm−1 for n in the above equation to calculate d.

d=12000.0 cm−1=5.00×10−4 cm(10−2 m1 cm)=5.00×10−6 m

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 449.2 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°449.2 nm(10−9 m1 nm)=11.1≈11

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 665.1 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°665.1 nm(10−9 m1 nm)=7.5≈8

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

Conclusion:

Therefore, the number of spectral lines is 19.

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

Answer to Problem 43P

The number of spectral lines is 19.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 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 2000.0 cm−1 for n in the above equation to calculate d.

d=12000.0 cm−1=5.00×10−4 cm(10−2 m1 cm)=5.00×10−6 m

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 449.2 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°449.2 nm(10−9 m1 nm)=11.1≈11

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

m=dsinθλ

Substitute 90.0° for θ, 5.00×10−6 m for d and 665.1 nm for λ in the above equation to calculate m.

m=(5.00×10−6 m)sin90.0°665.1 nm(10−9 m1 nm)=7.5≈8

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

Conclusion:

Therefore, the number of spectral lines is 19.

Answer 20

Ch. 25.1 - Prob. 25.1CP Ch. 25.1 - Path Difference for Destructive Interference...Ch. 25.2 - Prob. 25.2PP Ch. 25.3 - Prob. 25.3CP Ch. 25.3 - Prob. 25.3PP Ch. 25.4 - Prob. 25.4PP Ch. 25.4 - Prob. 25.5PP Ch. 25.5 - Prob. 25.5CP Ch. 25.5 - Prob. 25.6PP Ch. 25.6 - Prob. 25.7PP

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

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

Answer to Problem 43P

Explanation of Solution

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

Answer to Problem 43P

Explanation of Solution

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