Question

White light containing wavelengths from 400 nm to 750 nm falls on a grating with 7500 lines/cm. How wide is the first-order spectrum on a screen 2.20 m away?
___m

Answer 1

For $\theta$₁ for wavelength 400 nm, we have

Using an equation,

d sin $\theta$₁ = m $\lambda$₁                                                                          { eq.1 }

where, m = diffraction order = 1

$\lambda$₁ = wavelength of white light = 400 x 10^-9 m

d = diffraction grating = 1 / (7500 lines/cm)

inserting these values in above eq.

sin $\theta$₁ = (1) (400 x 10^-9 m) / [(0.01 m / 7500)]

sin $\theta$₁ = (400 x 10^-9 m) (7500) / (0.01 m)

$\theta$₁ = sin^-1 (0.3)

$\theta$₁ = 17.46⁰

For $\theta$₂ for wavelength 750 nm, we have

sin $\theta$₂ = (1) (750 x 10^-9 m) / [(0.01 m / 7500)]

sin $\theta$₂ = (750 x 10^-9 m) (7500) / (0.01 m)

$\theta$₂ = sin^-1 (0.562)

$\theta$₂ = 34.19⁰

width of the first-order spectrum on a screen which will be given as :

w = D (tan $\theta$₂ - tan $\theta$₁)                                                                    { eq.2 }

where, D = separation distance = 2.2 m

inserting the values in eq.2,

w = (2.2 m) [tan 34.19⁰ - tan 17.46⁰]

w = (2.2 m) [(0.6793) - (0.3145)]

w = (2.2 m) (0.3648)

w = 0.8 m