Question

A diffraction grating has 2605 lines per centimeter, and it produces a principal maximum at = 30.3°. The grating is used with light that contains all wavelengths between 410 and 660 nm. What are the two wavelengths of the incident light that could have produced this maximum? Give the longer wavelength as answer (a).

Answer 1

so m = 3 ,4 wavelength between 410 to 660 nm so 484.25 nm and 645.6 nm light produce principal maximum

Step-by-step explanation:

given data

grating =2605 lines per centimeter

angle = 30.3°

wavelengths between 410 and 660 nm

to find out

two wavelengths of the incident light that could have produced this maximum

solution

we know diffraction grating has 2605 lines / cm

so d = ( 1/ 2605 cm )

and we know equation

d sinθ = m× λ

so λ = d sinθ / m

θ = 30.3 so sin30.3 =

λ = (1/2605) sin30.3 / m

λ = (1937 nm ) / m

here put m = 1 , 2 , 3 , 4

if m = 1

λ = (1937 nm ) / 1 = 1937 nm

if m = 2

λ = (1937 nm ) / 2 = 968.5 nm

if m = 3

λ = (1937 nm ) / 3 = 645.6 nm

if m = 4

λ = (1937 nm ) / 4 = 484.25 nm

so that m = 3 ,4 wavelength between 410 to 660 nm so 484.25 nm and 645.6 nm light produce principal maximum

and longest wavelength is between 410 to 660 nm is 645.6 nm