The position of first bright fringe appears on either side of the central fringe.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 16, Problem 2SP

(a)

To determine

The position of first bright fringe appears on either side of the central fringe.

(a)

Expert Solution

Answer to Problem 2SP

The position of first bright fringe is 36 mm away from the center at each side.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 580 nm for λ, 1 for n, 1.3 m for D and 0.044 mm for d to find y.

y=(1)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=36×10−3 m=(36×10−3 m)(1 mm10−3 m)=36 mm

Thus, the position of first bright fringe is 36 mm away from the center at each side of the central maxima.

Conclusion:

Thus, position of first bright fringe is 36 mm away from the center at each side of the central fringe.

(b)

To determine

The position of second bright fringe on either side of the central fringe.

(b)

Expert Solution

Answer to Problem 2SP

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 2, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(2)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=72×10−3 m=(72×10−3 m)(1 mm10−3 m)=72 mm

Thus, the position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Conclusion:

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

(c)

To determine

The position of first dark bright fringe appeared on either side of the central fringe.

(c)

Expert Solution

Answer to Problem 2SP

The position of first dark fringe is 54 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of dark fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 1 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=54×10−3 m=(54×10−3 m)(1 mm10−3 m)=54 mm

Thus, the position of first dark fringe is 54 mm away from the center at each side.

Conclusion:

Thus, position of first dark fringe is 54 mm away from the center at each side.

(d)

To determine

Sketch the diffraction pattern showing the position of the seven central fringes and mark the position of the fringes.

(d)

Expert Solution

Answer to Problem 2SP

The diffraction pattern is given in figure 1.

Explanation of Solution

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the dark fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=90×10−3 m=(90×10−3 m)(1 mm10−3 m)=90 mm

Thus, the position of second dark fringe is 90 mm away from the center at each side.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 3 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(3)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=108×10−3 m=(108×10−3 m)(1 mm10−3 m)=108 mm

Thus, the position of third bright fringe is 108 mm away from the center at each side of the central fringe.

Following figure gives the diffraction pattern.

The Physics of Everyday Phenomena, Chapter 16, Problem 2SP

Figure 1

Here, the first, second and third order bright fringes will appear at distance of 36 mm, 72 mm and 108 mm from center respectively at each side of the central maxima. The dark fringes of order 1 and 2 will form distances 54 mm and 90 mm respectively.

Conclusion:

The diffraction pattern is given in figure 1.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

You are standing a distance x = 1.75 m away from this mirror. The object you are looking at is y = 0.29 m from the mirror. The angle of incidence is θ = 30°. What is the exact distance from you to the image?

For each of the actions depicted below, a magnet and/or metal loop moves with velocity v→ (v→ is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop. For the action depicted in (Figure 5), indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop). I know that the current is clockwise, I just dont understand why. Please fully explain why it's clockwise, Thank you

A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

Answer 2

Question

Chapter 16, Problem 2SP

(a)

To determine

The position of first bright fringe appears on either side of the central fringe.

(a)

Expert Solution

Answer to Problem 2SP

The position of first bright fringe is 36 mm away from the center at each side.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 580 nm for λ, 1 for n, 1.3 m for D and 0.044 mm for d to find y.

y=(1)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=36×10−3 m=(36×10−3 m)(1 mm10−3 m)=36 mm

Thus, the position of first bright fringe is 36 mm away from the center at each side of the central maxima.

Conclusion:

Thus, position of first bright fringe is 36 mm away from the center at each side of the central fringe.

(b)

To determine

The position of second bright fringe on either side of the central fringe.

(b)

Expert Solution

Answer to Problem 2SP

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 2, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(2)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=72×10−3 m=(72×10−3 m)(1 mm10−3 m)=72 mm

Thus, the position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Conclusion:

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

(c)

To determine

The position of first dark bright fringe appeared on either side of the central fringe.

(c)

Expert Solution

Answer to Problem 2SP

The position of first dark fringe is 54 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of dark fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 1 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=54×10−3 m=(54×10−3 m)(1 mm10−3 m)=54 mm

Thus, the position of first dark fringe is 54 mm away from the center at each side.

Conclusion:

Thus, position of first dark fringe is 54 mm away from the center at each side.

(d)

To determine

Sketch the diffraction pattern showing the position of the seven central fringes and mark the position of the fringes.

(d)

Expert Solution

Answer to Problem 2SP

The diffraction pattern is given in figure 1.

Explanation of Solution

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the dark fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=90×10−3 m=(90×10−3 m)(1 mm10−3 m)=90 mm

Thus, the position of second dark fringe is 90 mm away from the center at each side.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 3 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(3)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=108×10−3 m=(108×10−3 m)(1 mm10−3 m)=108 mm

Thus, the position of third bright fringe is 108 mm away from the center at each side of the central fringe.

Following figure gives the diffraction pattern.

The Physics of Everyday Phenomena, Chapter 16, Problem 2SP

Figure 1

Here, the first, second and third order bright fringes will appear at distance of 36 mm, 72 mm and 108 mm from center respectively at each side of the central maxima. The dark fringes of order 1 and 2 will form distances 54 mm and 90 mm respectively.

Conclusion:

The diffraction pattern is given in figure 1.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 16, Problem 2SP

(a)

To determine

The position of first bright fringe appears on either side of the central fringe.

(a)

Expert Solution

Answer to Problem 2SP

The position of first bright fringe is 36 mm away from the center at each side.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 580 nm for λ, 1 for n, 1.3 m for D and 0.044 mm for d to find y.

y=(1)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=36×10−3 m=(36×10−3 m)(1 mm10−3 m)=36 mm

Thus, the position of first bright fringe is 36 mm away from the center at each side of the central maxima.

Conclusion:

Thus, position of first bright fringe is 36 mm away from the center at each side of the central fringe.

(b)

To determine

The position of second bright fringe on either side of the central fringe.

(b)

Expert Solution

Answer to Problem 2SP

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 2, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(2)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=72×10−3 m=(72×10−3 m)(1 mm10−3 m)=72 mm

Thus, the position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Conclusion:

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

(c)

To determine

The position of first dark bright fringe appeared on either side of the central fringe.

(c)

Expert Solution

Answer to Problem 2SP

The position of first dark fringe is 54 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of dark fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 1 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=54×10−3 m=(54×10−3 m)(1 mm10−3 m)=54 mm

Thus, the position of first dark fringe is 54 mm away from the center at each side.

Conclusion:

Thus, position of first dark fringe is 54 mm away from the center at each side.

(d)

To determine

Sketch the diffraction pattern showing the position of the seven central fringes and mark the position of the fringes.

(d)

Expert Solution

Answer to Problem 2SP

The diffraction pattern is given in figure 1.

Explanation of Solution

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the dark fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=90×10−3 m=(90×10−3 m)(1 mm10−3 m)=90 mm

Thus, the position of second dark fringe is 90 mm away from the center at each side.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 3 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(3)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=108×10−3 m=(108×10−3 m)(1 mm10−3 m)=108 mm

Thus, the position of third bright fringe is 108 mm away from the center at each side of the central fringe.

Following figure gives the diffraction pattern.

The Physics of Everyday Phenomena, Chapter 16, Problem 2SP

Figure 1

Here, the first, second and third order bright fringes will appear at distance of 36 mm, 72 mm and 108 mm from center respectively at each side of the central maxima. The dark fringes of order 1 and 2 will form distances 54 mm and 90 mm respectively.

Conclusion:

The diffraction pattern is given in figure 1.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 16, Problem 2SP

(a)

To determine

The position of first bright fringe appears on either side of the central fringe.

(a)

Expert Solution

Answer to Problem 2SP

The position of first bright fringe is 36 mm away from the center at each side.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 580 nm for λ, 1 for n, 1.3 m for D and 0.044 mm for d to find y.

y=(1)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=36×10−3 m=(36×10−3 m)(1 mm10−3 m)=36 mm

Thus, the position of first bright fringe is 36 mm away from the center at each side of the central maxima.

Conclusion:

Thus, position of first bright fringe is 36 mm away from the center at each side of the central fringe.

(b)

To determine

The position of second bright fringe on either side of the central fringe.

(b)

Expert Solution

Answer to Problem 2SP

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 2, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(2)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=72×10−3 m=(72×10−3 m)(1 mm10−3 m)=72 mm

Thus, the position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Conclusion:

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

(c)

To determine

The position of first dark bright fringe appeared on either side of the central fringe.

(c)

Expert Solution

Answer to Problem 2SP

The position of first dark fringe is 54 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of dark fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 1 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=54×10−3 m=(54×10−3 m)(1 mm10−3 m)=54 mm

Thus, the position of first dark fringe is 54 mm away from the center at each side.

Conclusion:

Thus, position of first dark fringe is 54 mm away from the center at each side.

(d)

To determine

Sketch the diffraction pattern showing the position of the seven central fringes and mark the position of the fringes.

(d)

Expert Solution

Answer to Problem 2SP

The diffraction pattern is given in figure 1.

Explanation of Solution

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the dark fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=90×10−3 m=(90×10−3 m)(1 mm10−3 m)=90 mm

Thus, the position of second dark fringe is 90 mm away from the center at each side.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 3 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(3)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=108×10−3 m=(108×10−3 m)(1 mm10−3 m)=108 mm

Thus, the position of third bright fringe is 108 mm away from the center at each side of the central fringe.

Following figure gives the diffraction pattern.

The Physics of Everyday Phenomena, Chapter 16, Problem 2SP

Figure 1

Here, the first, second and third order bright fringes will appear at distance of 36 mm, 72 mm and 108 mm from center respectively at each side of the central maxima. The dark fringes of order 1 and 2 will form distances 54 mm and 90 mm respectively.

Conclusion:

The diffraction pattern is given in figure 1.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 16, Problem 2SP

(a)

To determine

The position of first bright fringe appears on either side of the central fringe.

(a)

Expert Solution

Answer to Problem 2SP

The position of first bright fringe is 36 mm away from the center at each side.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 580 nm for λ, 1 for n, 1.3 m for D and 0.044 mm for d to find y.

y=(1)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=36×10−3 m=(36×10−3 m)(1 mm10−3 m)=36 mm

Thus, the position of first bright fringe is 36 mm away from the center at each side of the central maxima.

Conclusion:

Thus, position of first bright fringe is 36 mm away from the center at each side of the central fringe.

(b)

To determine

The position of second bright fringe on either side of the central fringe.

(b)

Expert Solution

Answer to Problem 2SP

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 2, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(2)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=72×10−3 m=(72×10−3 m)(1 mm10−3 m)=72 mm

Thus, the position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Conclusion:

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

(c)

To determine

The position of first dark bright fringe appeared on either side of the central fringe.

(c)

Expert Solution

Answer to Problem 2SP

The position of first dark fringe is 54 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of dark fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 1 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=54×10−3 m=(54×10−3 m)(1 mm10−3 m)=54 mm

Thus, the position of first dark fringe is 54 mm away from the center at each side.

Conclusion:

Thus, position of first dark fringe is 54 mm away from the center at each side.

(d)

To determine

Sketch the diffraction pattern showing the position of the seven central fringes and mark the position of the fringes.

(d)

Expert Solution

Answer to Problem 2SP

The diffraction pattern is given in figure 1.

Explanation of Solution

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the dark fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=90×10−3 m=(90×10−3 m)(1 mm10−3 m)=90 mm

Thus, the position of second dark fringe is 90 mm away from the center at each side.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 3 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(3)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=108×10−3 m=(108×10−3 m)(1 mm10−3 m)=108 mm

Thus, the position of third bright fringe is 108 mm away from the center at each side of the central fringe.

Following figure gives the diffraction pattern.

The Physics of Everyday Phenomena, Chapter 16, Problem 2SP

Figure 1

Here, the first, second and third order bright fringes will appear at distance of 36 mm, 72 mm and 108 mm from center respectively at each side of the central maxima. The dark fringes of order 1 and 2 will form distances 54 mm and 90 mm respectively.

Conclusion:

The diffraction pattern is given in figure 1.

Answer 6

Chapter 16, Problem 2SP

(a)

To determine

The position of first bright fringe appears on either side of the central fringe.

(a)

Expert Solution

Answer to Problem 2SP

The position of first bright fringe is 36 mm away from the center at each side.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 580 nm for λ, 1 for n, 1.3 m for D and 0.044 mm for d to find y.

y=(1)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=36×10−3 m=(36×10−3 m)(1 mm10−3 m)=36 mm

Thus, the position of first bright fringe is 36 mm away from the center at each side of the central maxima.

Conclusion:

Thus, position of first bright fringe is 36 mm away from the center at each side of the central fringe.

Answer 7

Chapter 16, Problem 2SP

(a)

To determine

The position of first bright fringe appears on either side of the central fringe.

Answer 8

(a)

Expert Solution

Answer to Problem 2SP

The position of first bright fringe is 36 mm away from the center at each side.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 580 nm for λ, 1 for n, 1.3 m for D and 0.044 mm for d to find y.

y=(1)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=36×10−3 m=(36×10−3 m)(1 mm10−3 m)=36 mm

Thus, the position of first bright fringe is 36 mm away from the center at each side of the central maxima.

Conclusion:

Thus, position of first bright fringe is 36 mm away from the center at each side of the central fringe.

Answer 13

(b)

To determine

The position of second bright fringe on either side of the central fringe.

(b)

Expert Solution

Answer to Problem 2SP

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 2, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(2)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=72×10−3 m=(72×10−3 m)(1 mm10−3 m)=72 mm

Thus, the position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Conclusion:

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Answer 14

(b)

To determine

The position of second bright fringe on either side of the central fringe.

Answer 15

(b)

Expert Solution

Answer to Problem 2SP

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of bright fringe is 2, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(2)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=72×10−3 m=(72×10−3 m)(1 mm10−3 m)=72 mm

Thus, the position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Conclusion:

The position of second bright fringe is 72 mm away from the center at each side of the central fringe.

Answer 20

(c)

To determine

The position of first dark bright fringe appeared on either side of the central fringe.

(c)

Expert Solution

Answer to Problem 2SP

The position of first dark fringe is 54 mm away from the center at each side of the central fringe.

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of dark fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 1 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=54×10−3 m=(54×10−3 m)(1 mm10−3 m)=54 mm

Thus, the position of first dark fringe is 54 mm away from the center at each side.

Conclusion:

Thus, position of first dark fringe is 54 mm away from the center at each side.

Answer 21

(c)

To determine

The position of first dark bright fringe appeared on either side of the central fringe.

Answer 22

(c)

Expert Solution

Answer to Problem 2SP

The position of first dark fringe is 54 mm away from the center at each side of the central fringe.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: Wavelength of the light is 640 nm, order of dark fringe is 1, the distance of screen from slits is 1.8 m and spacing between adjacent slits is 0.032 mm.

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 1 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=54×10−3 m=(54×10−3 m)(1 mm10−3 m)=54 mm

Thus, the position of first dark fringe is 54 mm away from the center at each side.

Conclusion:

Thus, position of first dark fringe is 54 mm away from the center at each side.

Answer 27

(d)

To determine

Sketch the diffraction pattern showing the position of the seven central fringes and mark the position of the fringes.

(d)

Expert Solution

Answer to Problem 2SP

The diffraction pattern is given in figure 1.

Explanation of Solution

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the dark fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=90×10−3 m=(90×10−3 m)(1 mm10−3 m)=90 mm

Thus, the position of second dark fringe is 90 mm away from the center at each side.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 3 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(3)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=108×10−3 m=(108×10−3 m)(1 mm10−3 m)=108 mm

Thus, the position of third bright fringe is 108 mm away from the center at each side of the central fringe.

Following figure gives the diffraction pattern.

The Physics of Everyday Phenomena, Chapter 16, Problem 2SP

Figure 1

Here, the first, second and third order bright fringes will appear at distance of 36 mm, 72 mm and 108 mm from center respectively at each side of the central maxima. The dark fringes of order 1 and 2 will form distances 54 mm and 90 mm respectively.

Conclusion:

The diffraction pattern is given in figure 1.

Answer 28

(d)

To determine

Sketch the diffraction pattern showing the position of the seven central fringes and mark the position of the fringes.

Answer 29

(d)

Expert Solution

Answer to Problem 2SP

The diffraction pattern is given in figure 1.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write an expression for condition of minima.

y=(n+12)λDd

Here,

y is the distance of nth dark fringe from center

n is the order of the dark fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 2 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(1+12)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=90×10−3 m=(90×10−3 m)(1 mm10−3 m)=90 mm

Thus, the position of second dark fringe is 90 mm away from the center at each side.

Write an expression for condition of maxima.

y=nλDd

Here,

y is the distance of nth bright fringe from center

n is the order of the bright fringe

D is the screen’s distance from slit

λ is the wavelength

d is the slit separation

Substitute 640 nm for λ, 3 for n, 1.8 m for D and 0.032 mm for d to find y.

y=(3)((640 nm)(1 m10−9 nm))(1.8 m)(0.032 mm)(1 m103 mm)=108×10−3 m=(108×10−3 m)(1 mm10−3 m)=108 mm

Thus, the position of third bright fringe is 108 mm away from the center at each side of the central fringe.

Following figure gives the diffraction pattern.

The Physics of Everyday Phenomena, Chapter 16, Problem 2SP

Figure 1

Here, the first, second and third order bright fringes will appear at distance of 36 mm, 72 mm and 108 mm from center respectively at each side of the central maxima. The dark fringes of order 1 and 2 will form distances 54 mm and 90 mm respectively.