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Calculate minimum no. of lines in a grating which will resolve in the first order wavelengths are

$5890 A^o \ \ and \ \ 5896 A^o$


Mumbai university > FE > SEM 2 > Applied Physics II

Marks: 5M

Year: Dec 2014

1 Answer
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Given:-

m = 1

$λ_1 = 5890A^o$

$λ_2 = 5896A^o$

To find:- N

Solution:-

The Resolving power is given by

$\frac{λ}{dλ} = mN$

$dλ = |λ_1 - λ_2| = |5896 - 5890| = 6A^o$

$λ_2 \lt λ_1$ R.P for $λ_1$ will be less than $λ_2$

Hence, we calculate N for $λ_1$ as the same will resolve for $λ_2$

$\frac{λ_1}{dλ} = mN$

$N = \frac{5890}{6}$

$N = 981.67 ≈ 982$

$N = 982$

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