EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 33, Problem 27P

(a)

To determine

Thickness of the film.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The refractive index of film is 1.50 .

The missing wavelengths of spectrum are 360nm,450nm,602nm which means these are in destructive interface.

Formula used:

Write the expression for condition of destructive interference.

  2t=mλn

Here, t is the thickness of film, m is the order of spectrum, λ is the wavelength of light and n is the refractive index of film.

Rearrange the above equation for λ .

  λ=2ntm …… (1)

Rearrange the above equation for t .

  t=mλ2n …… (2)

Calculation:

Substitute 450nm for λ in equation (1).

  450nm=2ntm …… (3)

Substitute 360nm for λ and m+1 for m due to next missing wavelength in equation (1).

  360nm=2ntm+1 …… (4)

Divide equation (3) by equation (4).

  450nm360nm=m+1m54=m+1m

Solve for m .

  5m=4(m+1)m=4

Substitute 4 for m , 450nm for λ and 1.50 for n in equation (2).

  t=(4)( 450nm)2( 1.50)=600nm

Conclusion:

Thus, the thickness of film is 600nm .

(b)

To determine

The brightest wavelength in reflected interference pattern.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The refractive index of film is 1.50 .

The missing wavelengths of spectrum are 360nm,450nm,602nm which means these are in destructive interface.

Formula used:

Write the expression for condition of constructive interference.

  2t=(m+12)λn

Here, t is the thickness of film, m is the order of spectrum, λ is the wavelength of light and n is the refractive index of film.

Rearrange the above equation for λ .

  λ=2nt(m+12) …… (5)

Calculation:

Substitute 1.50 for n , 600nm for t in equation (1).

  λ=2(1.50)(600nm)(m+12)

  λ=1800nm(m+12) …… (6)

Substitute 0 for m in equation (6).

  λ=1800nm( 0+ 1 2 )=3600nm

Substitute 1 for m in equation (6).

  λ=1800nm( 1+ 1 2 )=( 1800nm)( 3 2 )=1200nm

Substitute 2 for m in equation (6).

  λ=1800nm( 2+ 1 2 )=1800nm( 5 2 )=720nm

Substitute 3 for m in equation (6).

  λ=1800nm( 3+ 1 2 )=1800nm( 7 2 )=514nm

Substitute 4 for m in equation (6).

  λ=1800nm( 4+ 1 2 )=1800nm( 9 2 )=400nm

Substitute 5 for m in equation (6).

  λ=1800nm( 5+ 1 2 )=1800nm( 11 2 )=327nm

Conclusion:

Thus, the wavelengths in visible spectrum are 720nm,514nm,400nm .

(c)

To determine

The wavelengths in visible spectrum when refractive index of glass is 1.60 .

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The refractive index of film is 1.60 .

The missing wavelengths of spectrum are 360nm,450nm,602nm which means these are in destructive interface.

Formula used:

Write the expression for condition of constructive interference.

  2t=(m+12)λn

Here, t is the thickness of film, m is the order of spectrum, λ is the wavelength of light and n is the refractive index of film.

Rearrange the above equation for λ .

  λ=2nt(m+12) …… (7)

Calculation:

Substitute 1.60 for n , 600nm for t in equation (7).

  λ=2(1.60)(600nm)(m+12)

  λ=1920nm(m+12) …… (8)

Substitute 0 for m in equation (8).

  λ=1920nm( 0+ 1 2 )=3840nm

Substitute 1 for m in equation (8).

  λ=1920nm( 1+ 1 2 )=( 1920nm)( 3 2 )=1280nm

Substitute 2 for m in equation (8).

  λ=1920nm( 2+ 1 2 )=1920nm( 5 2 )=768nm

Substitute 3 for m in equation (8).

  λ=1920nm( 3+ 1 2 )=1920nm( 7 2 )=548.5nm

Substitute 4 for m in equation (8).

  λ=1920nm( 4+ 1 2 )=1920nm( 9 2 )=426.7nm

Substitute 5 for m in equation (8).

  λ=1920nm( 5+ 1 2 )=1920nm( 11 2 )=349nm

Conclusion:

Thus, the wavelengths in visible spectrum are 548.5nm,426.7nm .

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Chapter 33 Solutions

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