General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 23, Problem 32E
To determine

Position of the first line.

Expert Solution & Answer
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Answer to Problem 32E

The position of the first line from the centre of the screen is 0.24m.

Explanation of Solution

Write the expression for diffraction grating.

  dsinθ=mλ        (I)

Here, λ is the wavelength of light, m is the order of interference, d is the distance between the slits and θ is the angle of the beam.

When θ is small sinθθ the above equation can be written as

  dθ=mλ        (II)

Write the expression for angle of the beam.

  θ=yy2+D2        (III)

Here, y is the position of the fringe and D is the distance between the source and screen.

Substitute equation (III) in equation (II)

  yy2+D2=mλdy2y2+D2=(mλd)2y2d2=(mλ)2(y2+D2)y2d2=m2λ2y2+D2m2λ2

  (y2d2m2λ2y2)=D2m2λ2y2(d2m2λ2)=D2m2λ2y2=D2m2λ2(d2m2λ2)

  y=(D2m2λ2(d2m2λ2))12        (IV)

Write the expression for space between the two adjacent lines.

  d=1N        (V)

Here, N is the number of lines per centimetre in the grating.

Conclusion:

Substitute 8000cm for N in equation (V) to find d

  d=18000cm=1.25×104cm

The distance between the two adjacent lines is 1.25×104m

Substitute 1.25×104cm for d, 0.5m for D, 550nm for y, 1 for m in equation (IV) to find y

  y=((0.5m)2(1)2(550×109m)2((1.25×104cm(0.01m1cm))2(1)2(550×109m)2))12=((0.25m2)(3.025×1013m2)(1.25×106m)2(3.025×1013m2))12=((7.5625×10-14m4)(1.5625×10-12m2)-(3.025×10-13m2))12=((7.5625×1014m4)(1.2601×1012m2))12=(0.060m2)12=0.24m

Therefore, the position of the first line from the centre of the screen is 0.24m.

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Chapter 23 Solutions

General Physics, 2nd Edition

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