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.

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