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 36P

(a)

To determine

The separation of the slits.

(a)

Expert Solution
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Explanation of Solution

Given:

The wavelength of the light used is 633nm .

The distance between the slit and screen is 12m .

The distance of first interference from the central maxima is 82cm .

Formula used:

The figure below shows the geometry of the given setup.

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 33, Problem 36P

Here L is the distance between the slit and screen, y1 is the distance of first interference from the central maxima, λ is the wavelength of the light used, d is the distance between the two slits, θ1 is the angle between first interference and the central maxima and θ is the path difference between the two waves.

Write the expression for two-slit interference maxima for d<<L .

  sinθ1λd

Rearrange the above expression for the distance between the two slits.

  dλsinθ1   ...... (1)

Write the expression for the triangle in the above figure with sides L and y1 .

  sinθ1=y1 ( L )2+ ( y 1 )2   ...... (2)

Calculation:

Substitute 12m for L and 82cm for y1 in equation (2).

  sinθ1=( 82cm) ( 12m( 10 2 cm m ) ) 2 + ( 82cm ) 2 =( 82cm)(1202.8cm)=0.06817

Substitute 633nm for λ and 0.06817 for sinθ1 in equation (1).

  d( 633nm)0.06817=9285.6nm=9285.6nm( 1μm 10 3 nm)9.3μm

Conclusion:

Thus, the separation of the slits is 9.3μm .

(b)

To determine

The possible number of interference fringes.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The wavelength of the light used is 633nm .

The distance between the slit and screen is 12m .

The distance of first interference from the central maxima is 82cm .

Formula used:

Write the expression for two-slit interference maxima for d<<L .

  sinθ1λd

Here L is the distance between the slit and screen, λ is the wavelength of the light used, θ1 is the angle between first interference and d is the distance between the two slits.

Rearrange the above expression for the distance between the two slits.

  dλsinθ1   ...... (1)

Write the expression for the triangle in the above figure with sides L and y1 .

  sinθ1=y1 ( L )2+ ( y 1 )2   ...... (2)

Here y1 is the distance of first interference from the central maxima.

Write the expression for the two slit interference maxima.

  dsinθ=mλ

where m is an integer.

Write the above expression for maximum number of interference fringes, for sinθ1 .

  d=mmaxλ

Here mmax is the maximum bright fringe.

Simplify above expression for mmax .

  mmax=dλ   ...... (3)

Write the expression for the maximum number of fringes in the central maximum.

  N=2mmax+1   ...... (4)

Here N is the number of fringes in the central maximum

Calculation:

Substitute 12m for L and 82cm for y1 in equation (2).

  sinθ1=( 82cm) ( 12m( 10 2 cm m ) ) 2 + ( 82cm ) 2 =( 82cm)(1202.8cm)=0.06817

Substitute 633nm for λ and 0.06817 for sinθ1 in equation (1).

  d( 633nm)0.06817=9285.6nm=9285.6nm( 1μm 10 3 nm)9.3μm

Substitute 9.3μm for d and 633nm for λ in equation (3).

  mmax=( 9.3μm)( 10 3 nm 1μm )633nm=9300633=14

Substitute 14 for mmax in equation (4).

  N=2(14)+1=28+1=29

Conclusion:

Thus, the possible number of interference fringes is 29 .

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Chapter 33 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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