Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 37, Problem 37.20P

Monochromatic light of wavelength λ is incident on a pair of slits separated by 2.40 × 10−4 m and forms an interference pattern on a screen placed 1.80 m from the slits. The first-order bright fringe is at a position ybright = 4.52 mm measured from the center of the central maximum. From this information, we wish to predict where the fringe for n = 50 would be located. (a) Assuming the fringes are laid out linearly along the screen, find the position of the n = 50 fringe by multiplying the position of the n = 1 fringe by 50.0. (b) Find the tangent of the angle the first-order bright fringe makes with respect to the line extending from the point midway between the slits to the center of the central maximum. (c) Using the result of part (b) and Equation 36.2, calculate the wavelength of the light. (d) Compute the angle for the 50th-order bright fringe from Equation 36.2. (e) Find the position of the 50th-order bright fringe on the screen from Equation 36.5. (f) Comment on the agreement between the answers to parts (a) and (e).

(a)

Expert Solution
Check Mark
To determine
The position of the 50th fringe.

Answer to Problem 37.20P

The position of the 50th fringe is 22.6cm .

Explanation of Solution

Given info: The separation of the slit is 2.40×104m , the distance between the screen and the slit is 1.80m and the position of the first-order bright fringe is ybright=4.52mm .

The formula to calculate the position of the 50th fringe is,

y=50(ybright)m=1

Here,

ybright is the position of the first-order bright fringe.

Substitute 4.52mm for ybright in above equation to find the value of y .

y=50(4.52mm)=22.6cm

Conclusion:

Therefore, the position of the 50th fringe is 22.6cm .

(b)

Expert Solution
Check Mark
To determine
The tangent of the angle of the first-order bright fringe with respect to the point midway between the slits to the center of the central maximum.

Answer to Problem 37.20P

The tangent of the angle of the first-order bright fringe with respect to the point midway between the slits to the center of the central maximum is 2.51×103 .

Explanation of Solution

Given info: The separation of the slit is 2.40×104m , the distance between the screen and the slit is 1.80m and the position of the first-order bright fringe is ybright=4.52mm .

The formula to calculate the tangent of the angle is,

tanθ1=(ybright)m=1L

Here,

L is the distance between the screen and the slit.

Substitute 4.52mm for (ybright)m=1 and 1.80m for L in the above equation.

tanθ1=4.52mm×103m1m1.80m=2.51×103

Conclusion:

Therefore, the tangent of the angle of the first-order bright fringe with respect to the point midway between the slits to the center of the central maximum is 2.51×103 .

(c)

Expert Solution
Check Mark
To determine
The wavelength of the light.

Answer to Problem 37.20P

The wavelength of the light is 6.03×107m .

Explanation of Solution

Given info: The separation of the slit is 2.40×104m , the distance between the screen and the slit is 1.80m and the position of the first-order bright fringe is ybright=4.52mm .

The formula to calculate the tangent of the angle is,

tanθ1=(ybright)m=1L

Substitute 4.52mm for (ybright)m=1 and 1.80m for L in the above equation.

tanθ1=4.52mm×103m1m1.80mtanθ1=2.51×103θ1=0.144° ]

The formula to calculate the wavelength is,

mλ=dsinθ1

Here,

m is the number of the maxima.

λ is the wavelength of the light.

Substitute 1 for m , 2.40×104m for d and 0.144° for θ1 in above equation.

λ=(2.40×104m)sin(0.144°)=6.03×107m

Conclusion:

Therefore, the wavelength of the light is 6.03×107m .

(d)

Expert Solution
Check Mark
To determine
The angle of the 50th order-bright fringe.

Answer to Problem 37.20P

The angle of the 50th order-bright fringe is 7.21° .

Explanation of Solution

Given info: The separation of the slit is 2.40×104m , the distance between the screen and the slit is 1.80m and the position of the first-order bright fringe is ybright=4.52mm .

The formula to calculate the angle of the 50th order-bright fringe is,

θ50=sin1(50sinθ1)

Substitute 0.144° for θ1 in the above equation.

θ50=sin1(50sin(0.144°))=7.21°

Conclusion:

Therefore, the angle of the 50th order-bright fringe is 7.21° .

(e)

Expert Solution
Check Mark
To determine
The position of the 50th order-bright fringe.

Answer to Problem 37.20P

The position of the 50th order-bright fringe is 2.28cm .

Explanation of Solution

Given info: The separation of the slit is 2.40×104m , the distance between the screen and the slit is 1.80m and the position of the first-order bright fringe is ybright=4.52mm .

The formula to calculate the position of the 50th order-bright fringe is,

y50=Ltanθ50

Substitute 1.80m for L and 7.21° for θ50 in above equation to find the value of y50 .

y50=(1.80m)tan(7.21°)=2.28×102m×102cm1m=2.28cm

Conclusion:

Therefore, the position of the 50th order-bright fringe is 2.28cm .

(f)

Expert Solution
Check Mark
To determine
The comment on the agreement between the answers to parts (a) and part (e) .

Answer to Problem 37.20P

The answer to part (a) and part (e) are very close bet not equal as the fringes are not laid linear.

Explanation of Solution

Given info: The separation of the slit is 2.40×104m , the distance between the screen and the slit is 1.80m and the position of the first-order bright fringe is ybright=4.52mm .

The difference in the position of the 50th order-bright fringe is different as calculated in part (a) and part (e) so it can be deduced from the results that the fringes are not laid out linearly on the screen as assumed in part (a) and the nonlinearity is evident for very large angles.

Conclusion:

Therefore, the answer to part (a) and part (e) are very close bet not equal as the fringes are not laid linear.

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Chapter 37 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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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