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14 May, 13:51

A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstration. She shines a beam of white light through a diffraction grating that has 600 lines per mm, projecting a pattern on a screen 2.9 m behind the grating. How much distance separates the end of the m = 1 spectrum and the start of the m = 2 spectrum? I already determined that the m=1 spectrum is 0.63 meters wide.

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  1. 14 May, 14:10
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    Spectrum width = 0.784m

    Explanation:

    We are given that;

    λ1 = 400 nm = 400 x 10^ (-9)

    λ2 = 700 nm = 700 x 10^ (-9)

    Slit width; d = 10^ (-3) / 700 = 1.428 x 10^ (-6) m

    Distance of screen from light source; D = 2.9m

    Now, in slit Experiment, we know that;

    d•sinθ = nλ

    where;

    θ = angle between the path and a line from the slits to the screen.

    Thus,

    1.428 x 10^ (-6) x sinθ = 1 x 400 x 10^ (-9)

    sinθ = [1 x 400 x 10^ (-9) ] / (1.428 x 10^ (-6))

    sinθ = 0.28011

    θ = sin^ (-1) 0.28011

    θ = 16.27°

    Now, tan θ = y/D

    where y is the position of First maxima corresponding

    Thus,

    y1 = D. tanθ

    y1 = 2.9 x tan16.27

    y1 = 0.846m

    Now, let's do the same for λ2;

    1.428 x 10^ (-6) x sinθ' = 1 x 700 x 10^ (-9)

    sinθ' = [1 x 700 x 10^ (-9) ] / (1.428 x 10^ (-6))

    sinθ' = 0.4901

    θ = sin^ (-1) 0.4901

    θ' = 29.35°

    Now, tan θ' = y'/D

    where y is the position of First maxima corresponding

    Thus,

    y'1 = D. tanθ

    y'1 = 2.9 x tan29.35

    y'1 = 1.63 m

    Spectrum width = y'1 - y1 = 1.63 - 0.846 = 0.784m
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