0
6.9kviews
Find the maximum value of RP a grating 3 cm wide & having 7000 lines/cm can have if a light of λ = 5000 ˚A is used.

Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

1 Answer
2
1.6kviews

Width W = 3 cm, N = 7000 lines/cm, λ = 5000 A°

grating element (a+b)=1N=17000

RP is maximum when n is maximum

N is maximum when sin θ =1

(a+b) sin θ = n λ

n_{max} = \frac{1}{Nλ} = \frac{1}{ 7000 \times 5000 \times 10^{-8}} = 2.8

n cannot be 3 as sin θ exceeds one

n_{max} = 2

RP = n N

RP = 2x7000 x 3 as grating is 3 cm wide

RP = 42000

Please log in to add an answer.