Find the maximum value of RP a grating 3 cm wide & having 7000 lines/cm can have if a light of $\lambda$ = 5000 $\mathring{A}$ is used.

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written 7.9 years ago by

neeta.vanage • 880

modified 3.9 years ago by

pedsangini276 • 4.8k

Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

engineering physics

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1 Answer

1.6kviews

written 7.8 years ago by

neeta.vanage • 880

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

grating element $ (a+b) = \frac{1}{N} = \frac{1}{7000} $

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

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