Respuesta :
Answer:
n = 3 total orders seen
n = 1 , Q_1 = 15 degrees
n = 2 , Q_2 = 31 degrees
n = 3 , Q_3 = 50 degrees
Explanation:
Given:
- Number of grating lines per mm N = 500 line/mm
- Wavelength of the light λ = 510 nm
Find:
How many diffraction orders are seen, and what is the angle of each?
Solution:
- We know that the maximum angle of diffraction Q_m of the furthest bright fringe < Q = 90 degrees.
- We need to compute the nth bright fringe for which is approximated to 90 degrees.
- The angle of nth bright fringe is given by:
sin(Q_m) = n*λ*N
- Approximating Q_m ≈ 90 degrees.
sin (90) = n*λ*N
n = sin (90) / λ*N
n = 1 / ((510*10^-6*)500
n = 3.9 orders
- Since, we knew that Q_m < 90 degrees, we will choose n = 3 as the maximum number of orders. That means 3 fringes above and 3 fringes below the central order are observed.
- So for n = 1
sin(Q_n) = n*λ*N
sin(Q_1) = 1*(510*10^-9)*(500,000)
Q_1 = sin^-1(0.255)
Q_1 = 15 degrees
- So for n = 2
sin(Q_n) = n*λ*N
sin(Q_2) = 2*(510*10^-9)*(500,000)
Q_2 = sin^-1(0.51)
Q_2 = 31 degrees
- So for n = 3
sin(Q_n) = n*λ*N
sin(Q_3) = 3*(510*10^-9)*(500,000)
Q_3 = sin^-1(0.765)
Q_3 = 50 degrees
