Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 33, Problem 67P
To determine

The order of spectrum observe in transmitted light in range of 400nm to 700nm .

Expert Solution & Answer
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Explanation of Solution

Given:

The spacing of diffraction grating has 4800 lines per centimeter.

The range of wavelength of transmitted light is 400nm to 700nm .

Formula used:

Write the formula for slit separation.

  d=1N   ...... (1)

Here, d is the distance between two slits and N is the number of slits per centimeter.

Write the condition of maximum interface.

  dsinθ=mλ

Here, d is the slit separation, θ is angle at which first spectrum expected, m is the order of spectrum and λ is the wavelength of line.

Rearrange the above equationfor m .

  m=dsinθλ   ...... (2)

Rearrange the equation.

  sinθ=mλd

Rearrange the above equation for θ .

  θ=sin1(mλd)   ...... (3)

Calculation:

Substitute 4800 per centimeter for N in equation (1).

  d=1cm4800=0.00021cm

Substitute 1 for sinθ (maximum value of sinθ ), 0.00021cm for d and 400nm for λ in equation (2).

  m=( 0.00021cm)(1)400nm=( 0.00021cm)( 1m 100cm )( 400nm)( 1m 10 9 nm )5

Substitute 1 for sinθ (maximum value of sinθ ), 0.00021cm for d and 700nm for λ in equation (2).

  m=( 0.00021cm)(1)700nm=( 0.00021cm)( 1m 100cm )( 700nm)( 1m 10 9 nm )2

The order of spectrum in range 400nm to 700nm is 3(52) .

Substitute 3 for m , 400nm for λ and 0.00021cm for d in equation (3).

  θ=sin1( ( 3 )( 400nm ) ( 0.00021cm ))=sin1( ( 3 )( 400nm )( 1m 10 9 nm ) ( 0.00021cm )( 1m 100cm ))=sin1(0.576)=35.17°

Substitute 4 for m , 400nm for λ and 0.00021cm for d in equation (3).

  θ=sin1( ( 4 )( 400nm ) ( 0.00021cm ))=sin1( ( 4 )( 400nm )( 1m 10 9 nm ) ( 0.00021cm )( 1m 100cm ))=sin1(0.768)=50.174°

Substitute 5 for m , 400nm for λ and 0.00021cm for d in equation 3.

  θ=sin1( ( 5 )( 400nm ) ( 0.00021cm ))=sin1( ( 5 )( 400nm )( 1m 10 9 nm ) ( 0.00021cm )( 1m 100cm ))=sin1(0.960)=73.739°

Angle for λ=700nm are out of range. So these order overlap with each other and the overlapping regions will be for λ>400nm .

Conclusion:

Thus, the order of spectrum range from 400nm to 700nm is 5 and there will be overlap of long wavelengths of second order spectrum with the short wavelengths of third order spectrum.

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How to rized rigu o ast 15s/AUT. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. The spacing of ruled lines on a diffraction grating is 1830 mm. The grating is illuminated at normal incidence with a parallel beam of white light in the 400 nm to 700 mm wavelength band. The second order spectrum and the third order spectrum overlap. The angular width of the overlap is closest to: A) 6.9° B) 8.9° C) 4.9° D) 5.9° 7 E) 7.9°
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