Light of wavelength 450 nm produces a first-order maximum at 27$^{\circ}$when viewed through a grating.At what angle would a first order maximum occur for a wavelength of 600 nm viewed through this grating?
A) 54$^{\circ}$
B) 40$^{\circ}$
C) 36$^{\circ}$
D) 37$^{\circ}$
E) 14$^{\circ}$

Question

Quizplus · Accepted Answer

Using the grating equation,

d*sin(θ) = mλ

where d is the grating spacing, θ is the angle of diffraction, m is the order of the maximum, and λ is the wavelength of light.

For the first-order maximum for the given wavelength (450 nm),

d*sin(27°) = 1*450 nm

Rearranging this equation,

sin(θ) = (450 nm)/(d) * sin(27°)

For the first-order maximum for a wavelength of 600 nm,

sin(θ) = (600 nm)/(d) * sin(θ')

where θ' is the angle we want to find.

Setting the two sin equations equal to each other,

(450 nm)/(d) * sin(27°) = (600 nm)/(d) * sin(θ')

Simplifying,

sin(θ') = sin(27°) * (450 nm)/(600 nm)

Taking the inverse sine,

θ' = sin⁻¹(sin(27°) * (450 nm)/(600 nm))

θ' = 37.3°

Rounding to one decimal place, the answer is D, 37°.

Light of Wavelength 450 Nm Produces a First-Order Maximum at 27