Question

A diffraction grating gives a second-order maximum at as angle of 31° for violet light (λ = 4.0 × 102 nm). If the diffraction grating is 1.0 cm in width, how many lines are on this diffraction grating? [___/3]

Answer 1

for second order maximum we can say

`a sin\theta = (5\lambda)/(2)`

here we will have

`a= (0.01)/(N)`

now we will have

`(0.01)/(N) sin31 = (5*400*10^(-9))/(2)`

`N = (0.01 * sin31)/(5*200*10^(-9))`

`N = 5150`

so the number of lines are 5150.