Question

How many orders will be visible if the wavelength of the incident radiation is 550 nm and the number of lines per inch on grating is1500?​

Answer 1

dsinθ = mλ . when λ = 90o m = -mmax and m = mmax

where d is the diffraction grating line spacing, λ is the wavelength of the light source, θ is the

angle form the grating to the screen, and m = -mmax .... 0 ... mmax (integers)

We know that b + d  the maximum value of θ is 900, that is sinθ = 1. Therefore, the number of orders visible with the granting is given as b + d = nλ

the grating is 2500 lines per inch.

m = \frac{b + d}{λ}  

λ

b+d

​

 

here b + d = 2500 lines per inch to 2.54 / 2500 to per cm

and λ = 5000\dot{A}  

A

˙

 = 5000 * 10-8cm

m = \frac{2.54}{2500 * 4800 * 10^{-8}}  

2500∗4800∗10  

−8

 

2.54

​

 = 21.6

m > 21

21 orders will be visible

Answer 2

Given: the wavelength of the incident radiation is 550 nm and the number of lines per inch on the grating is 1500

To find: Number of orders visible

Solution: we know that the angular position of principal maxima can be calculated through the formula

(a+b)sinθ = mλ

here (a+b) is the distance between two consecutive slits called gradient element

m is order of principal maxima on which we get spectral lines

the maximum value of θ can be 90°

we need to find number of orders visible that will be m

m= (a+d)/λ

here we are given (a+d) = 2.54/ 1500

m = 2.54/1500× 550×10^-7

m = 3

Therefore, Number of orders visible will be 3.