Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 27, Problem 18P

(a)

To determine

The position of the 50th fringe.

(a)

Expert Solution
Check Mark

Answer to Problem 18P

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)

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.

(b)

Expert Solution
Check Mark

Answer to Problem 18P

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)

To determine

The wavelength of the light.

(c)

Expert Solution
Check Mark

Answer to Problem 18P

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)

To determine

The angle of the 50th order-bright fringe.

(d)

Expert Solution
Check Mark

Answer to Problem 18P

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)

To determine

The position of the 50th order-bright fringe.

(e)

Expert Solution
Check Mark

Answer to Problem 18P

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)

To determine

The comment on the agreement between the answers to parts (a) and part (e) .

(f)

Expert Solution
Check Mark

Answer to Problem 18P

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 27 Solutions

Principles of Physics: A Calculus-Based Text

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