Question

White light (400–700 nm) is incident on a 600 line/mm diffraction grating. What is the width of the first-order rainbow on a screen 2.0 m behind the grating?

Answer 1

Answer:

Δy=0.431m

Explanation:

Diffraction grating with split space d,to find the fringe position ym,we must to find the angle from

dSinα=mλ

A grating with N slits or lines per mm has slit spacing of

d=1/N

d=(1/600mm)

d=1.67×10⁻³mm

For 400nm wavelength:

α=Sin⁻¹(mλ/d)

$\alpha =Sin^{-1}(\frac{400*10^{-9} }{1.67*10^{-6}} )\\ \alpha =13.910^{o}$

And the position of first order lowest wavelength fringe on the screen is:

$y_{1}=Ltan\alpha_{1}\\y_{1}=2tan(13.910)\\ y_{1}=0.49445m$

For 700nm wavelength:

α=Sin⁻¹(mλ/d)

$\alpha =Sin^{-1}(\frac{700*10^{-9} }{1.67*10^{-6}} )\\ \alpha_{2} =24.83^{o}$

And the position of first order highest wavelength fringe on the screen is:

$y_{2}=Ltan\alpha_{2}\\y_{2}=2tan(24.83)\\ y_{2}=0.925595m$

The difference between the first order lowest and highest wavelength fringe is

Δy=(0.925595 - 0.49445)m

Δy=0.431m