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Chapter 35, Problem 14PQ

(a)

To determine

The linear distance on the screen between the central maximum and the n=2 minimum.

(a)

Expert Solution
Check Mark

Answer to Problem 14PQ

Therefore the linear distance on the screen between the central maximum and the n=2 minimum is 4.75cm.

Explanation of Solution

Write the expression for the path difference for bright fringes in young’s double slit experiment.

dsinθn=(n+12)λ (I)

Here, λ is the wavelength of microwaves, d is the separation between two slits and n is the integer.

Calculate the angle made by a particular interference fringe on the screen from the central maximum.

tanθn=ynx (II)

Here, yn is the distance of the particular fringe from the central maximum, θn is the angle made by the nth  fringe and x is the distance from slits to the screen.

For small angle approximation tanθsinθ.

Substitute (n+12)λd for tanθ in equation (II).

ynx=(n+12)λdyn=(n+12)λxd (III)

Conclusion:

Substitute 50.0μm for d, 633nm for λ, 2 for n and 1.50m for x in equation (III) to calculate y2.

y2=(2+12)×(633nm)(109m1nm)(1.50m)50.0μm(106m1μm)=(52)×(633nm)(109m1nm)(1.50m)50.0μm(106m1μm)=4.75×102m(102cm1m)=4.75cm

Therefore the linear distance on the screen between the central maximum and the n=2 minimum is 4.75cm.

(b)

To determine

The linear distance on the screen between the central maximum and the n=20 minimum.

(b)

Expert Solution
Check Mark

Answer to Problem 14PQ

The linear distance on the screen between the central maximum and the n=20 maximum is 40.2cm.

Explanation of Solution

Write the expression for the path difference for dark fringes in young’s double slit experiment.

dsinθn=(n+12)λsinθn=(n+12)λdθn=sin1((n+12)λd) (IV)

Here, λ is the wavelength of microwaves, d is the separation between two slits and n is the integer.

Consider the equation (II)

tanθn=ynxyn=xtanθn (V)

Here, yn is the distance of the particular fringe from the central maximum, θn is the angle made by the nth  fringe and x is the distance from slits to the screen.

Conclusion:

Substitute 633nm for λ, 20 for n and 50.0μm for d in equation (IV) to calculate θ20.

θ20=sin1((20+12)(633nm)(109m1nm)50.0μm(106m1μm))=sin1((412)(633nm)(109m1nm)50.0μm(106m1μm))=15.0°

Substitute 1.50m for x, 20 for n and 15.0° for θ20 in equation (V) to calculate y20.

y20=(1.50m)tan(15.0°)=0.402m(102cm1m)=40.2cm

Therefore, the linear distance on the screen between the central maximum and the n=20 maximum is 40.2cm.

(c)

To determine

Whether the minima are evenly spaced.

(c)

Expert Solution
Check Mark

Answer to Problem 14PQ

No, the minima are not evenly spaced.

Explanation of Solution

No, the minima are not evenly spaced. The fringes with small angles have equal separation but when the integer n increases the separation between the subsequent minima increases. That is why the 20th minima will have more angular separation than the 2nd minima.

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Chapter 35 Solutions

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term

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