EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100581557
Author: Jewett
Publisher: Cengage Learning US
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Chapter 37, Problem 37.26P

Monochromatic coherent light of amplitude E0 and angular frequency ω passes through three parallel slits, each separated by a distance d from its neighbor. (a) Show that the time-averaged intensity as a function of the angle θ is

I ( θ ) = I max [ 1 + 2 cos ( 2 π d sin θ λ ) ] 2

(b) Explain how this expression describes both the primary and the secondary maxima. (c) Determine the ratio of the intensities of the primary and secondary maxima. Hint: See Problem 16.

(a)

Expert Solution
Check Mark
To determine

To show: The time averaged intensity as a function of angle θ is I(θ)=Imax[1+2cos(2πdsinθλ)]2 .

Answer to Problem 37.26P

The time averaged intensity as a function of angle θ is I(θ)=Imax[1+2cos(2πdsinθλ)]2 .

Explanation of Solution

Given info: The amplitude of monochromatic light is E0 and angular frequency is ω .

The amplitude of the monochromatic light is,

E1=E0sinωtE2=E0(sinωt+ϕ)E3=E0(sinωt+2ϕ)

Here,

ω is the angular frequency.

ϕ is the phase difference whose value is 2πdsin(θ)λ .

The total amplitude of monochromatic light is,

E=E1+E2+E3

Substitute E0sinωt for E1 , E0(sinωt+ϕ) for E2 , E0(sinωt+2ϕ) for E3 in the above formula as,

E=E0sinωt+E0(sinωt+ϕ)+E0(sinωt+2ϕ)

Apply the trigonometric identity to the above expression as,

E=2E0cos(ϕ2)sin(2ωt+ϕ2) +

Apply the trigonometric identity to the above expression as,

E=2E0cos(ϕ2)sin(2ωt+ϕ2)+E0sin(ωt+ϕ)

Add the above result to E2 to get the resultant field as,

E=2E0cos(ϕ2)sin(2ωt+ϕ2)+E0sin(ωt+ϕ)

As it is known that intensity of monochromatic light is directly proportional to the the square of the electric field that is,

IE2

Here,

I is the intensity.

E is the electric field.

The resultant field is obtained by square the above expression as,

E2=[[2E0cos(ϕ2)sin(2ωt+ϕ2)]+[E0sin(ωt+ϕ)]]2AB=4E20cos2(ϕ2)sin2(2ωt+ϕ2)+E02sin2(ωt+ϕ)+4E0cos(ϕ2)sin(2ωt+ϕ2)E0sin(ωt+ϕ)=E20(1+2cosϕ)2

Substitute 2πdsin(θ)λ for ϕ in the above expression as,

I(θ)=Imax[1+2cos(2πdsinθλ)]2

Conclusion:

Thus, the time averaged intensity as a function of angle θ is I(θ)=Imax[1+2cos(2πdsinθλ)]2 .

(b)

Expert Solution
Check Mark
To determine
The way in which the expression describes both the primary and secondary maxima.

Answer to Problem 37.26P

The expression describes both the primary and secondary maxima.

Explanation of Solution

Given info: The amplitude of monochromatic light is E0 and angular frequency is ω .

From the above expression obtained in part (a), the minimum interference is obtained when cosϕ is 12 and maximum interference is obtained when cosϕ is 1.

Conclusion:

Thus, the expression describes both the primary and secondary maxima.

(c)

Expert Solution
Check Mark
To determine
The ratio of the intensities of the primary and secondary maxima.

Answer to Problem 37.26P

The ratio of the intensities of the primary and secondary maxima is 19

Explanation of Solution

Given info: The amplitude of monochromatic light is E0 and angular frequency is ω .

Consider the amplitude of the monochromatic light is,

E1=E0sinωtE2=E0(sinωt+ϕ)E3=E0(sinωt+2ϕ)

The total amplitude of monochromatic light is,

E=E1+E2+E3

Substitute E0sinωt for E1 , E0(sinωt+ϕ) for E2 , E0(sinωt+2ϕ) for E3 in the above formula as,

E=E0sinωt+E0(sinωt+ϕ)+E0(sinωt+2ϕ)

Apply the trigonometric identity to the above expression as,

E=2E0cos(ϕ2)sin(2ωt+ϕ2) +

Apply the trigonometric identity to the above expression as,

E=2E0cos(ϕ2)sin(2ωt+ϕ2)+E0sin(ωt+ϕ)

Add the above result to E2 to get the resultant field as,

E=2E0cos(ϕ2)sin(2ωt+ϕ2)+E0sin(ωt+ϕ)

As it is known that intensity of monochromatic light is directly proportional to the  square of the electric field that is,

IE2

Here,

I is the intensity.

E is the electric field.

The resultant field is obtained by square the above expression as,

E2=[[2E0cos(ϕ2)sin(2ωt+ϕ2)]+[E0sin(ωt+ϕ)]]2=4E20cos2(ϕ2)sin2(2ωt+ϕ2)+E02sin2(ωt+ϕ)+4E0cos(ϕ2)sin(2ωt+ϕ2)E0sin(ωt+ϕ)=E20(1+2cosϕ)2

Substitute 2πdsin(θ)λ for ϕ in the above expression as,

I(θ)=Imax[1+2cos(2πdsinθλ)]2

The expression for the intensity for primary maxima is,

I(θ)=Imax[1+2cos(2πdsinθλ)]2

The expression for the intensity for secondary maxima is,

I(θ)=9Imax[1+2cos(2πdsinθλ)]2

Take the ratio of the above two expression as,

I(θ)priI(θ)sec=19

Conclusion:

Therefore, the ratio of the intensities of the primary and secondary maxima is 19 .

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Chapter 37 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 37 - Suppose you perform Youngs double-slit experiment...Ch. 37 - A plane monochromatic light wave is incident on a...Ch. 37 - A film of' oil on a puddle in a parking lot shows...Ch. 37 - Prob. 37.1CQCh. 37 - Prob. 37.2CQCh. 37 - Explain why two flashlights held close together do...Ch. 37 - A lens with outer radius of curvature R and index...Ch. 37 - Consider a dark fringe in a double-slit...Ch. 37 - Prob. 37.6CQCh. 37 - What is the necessary condition on the path length...Ch. 37 - In a laboratory accident, you spill two liquids...Ch. 37 - A theatrical smoke machine fills the space bet...Ch. 37 - Two slits are separated by 0.320 mm. A beam of...Ch. 37 - Light of wavelength 530 nm illuminates a pair of...Ch. 37 - A laser beam is incident on two slits with a...Ch. 37 - A Youngs interference experiment is performed with...Ch. 37 - Youngs double-slit experiment is performed with...Ch. 37 - Why is the following situation impossible? Two...Ch. 37 - Light of wavelength 620 nm falls on a double slit,...Ch. 37 - In a Youngs double-slit experiment, two parallel...Ch. 37 - pair of narrow, parallel slits separated by 0.250...Ch. 37 - Light with wavelength 442 nm passes through a...Ch. 37 - The two speakers of a boom box are 35.0 cm apart....Ch. 37 - Prob. 37.12PCh. 37 - Two radio antennas separated by d = 300 in as...Ch. 37 - A riverside warehouse has several small doors...Ch. 37 - A student holds a laser that emits light of...Ch. 37 - A student holds a laser that emits light of...Ch. 37 - Radio waves of wavelength 125 m from a galaxy...Ch. 37 - In Figure P36.10 (not to scale), let L = 1.20 m...Ch. 37 - Coherent light rays of wavelength strike a pair...Ch. 37 - Monochromatic light of wavelength is incident on...Ch. 37 - In the double-slit arrangement of Figure P36.13, d...Ch. 37 - Youngs double-slit experiment underlies the...Ch. 37 - Two slits are separated by 0.180 mm. An...Ch. 37 - Prob. 37.24PCh. 37 - In Figure P37.18, let L = 120 cm and d = 0.250 cm....Ch. 37 - Monochromatic coherent light of amplitude E0 and...Ch. 37 - The intensity on the screen at a certain point in...Ch. 37 - Green light ( = 546 nm) illuminates a pair of...Ch. 37 - Two narrow, parallel slits separated by 0.850 mm...Ch. 37 - A soap bubble (n = 1.33) floating in air has the...Ch. 37 - A thin film of oil (n = 1.25) is located on...Ch. 37 - A material having an index of refraction of 1.30...Ch. 37 - Prob. 37.33PCh. 37 - A film of MgF2 (n = 1.38) having thickness 1.00 ...Ch. 37 - A beam of 580-nm light passes through two closely...Ch. 37 - An oil film (n = 1.45) floating on water is...Ch. 37 - An air wedge is formed between two glass plates...Ch. 37 - Astronomers observe the chromosphere of the Sun...Ch. 37 - When a liquid is introduced into the air space...Ch. 37 - A lens made of glass (ng = 1.52) is coated with a...Ch. 37 - Two glass plates 10.0 cm long are in contact at...Ch. 37 - Mirror M1 in Figure 36.13 is moved through a...Ch. 37 - Prob. 37.43PCh. 37 - One leg of a Michelson interferometer contains an...Ch. 37 - Radio transmitter A operating at 60.0 MHz is 10.0...Ch. 37 - A room is 6.0 m long and 3.0 m wide. At the front...Ch. 37 - In an experiment similar to that of Example 36.1,...Ch. 37 - In the What If? section of Example 36.2, it was...Ch. 37 - An investigator finds a fiber at a crime scene...Ch. 37 - Raise your hand and hold it flat. Think of the...Ch. 37 - Two coherent waves, coming from sources at...Ch. 37 - In a Youngs interference experiment, the two slits...Ch. 37 - In a Youngs double-slit experiment using light of...Ch. 37 - Review. A flat piece of glass is held stationary...Ch. 37 - A certain grade of crude oil has an index of...Ch. 37 - The waves from a radio station can reach a home...Ch. 37 - Interference effects are produced at point P on a...Ch. 37 - Measurements are made of the intensity...Ch. 37 - Many cells are transparent anti colorless....Ch. 37 - Consider the double-slit arrangement shown in...Ch. 37 - Figure P36.35 shows a radio-wave transmitter and a...Ch. 37 - Figure P36.35 shows a radio-wave transmitter and a...Ch. 37 - In a Newtons-rings experiment, a plano-convex...Ch. 37 - Why is the following situation impossible? A piece...Ch. 37 - A plano-concave lens having index of refraction...Ch. 37 - A plano-convex lens has index of refraction n. The...Ch. 37 - Interference fringes are produced using Lloyds...Ch. 37 - Prob. 37.68APCh. 37 - Astronomers observe a 60.0-MHz radio source both...Ch. 37 - Figure CQ37.2 shows an unbroken soap film in a...Ch. 37 - Our discussion of the techniques for determining...Ch. 37 - The condition for constructive interference by...Ch. 37 - Both sides of a uniform film that has index of...Ch. 37 - Prob. 37.74CPCh. 37 - Monochromatic light of wavelength 620 nm passes...Ch. 37 - Prob. 37.76CP
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