Question

When light of a wavelength lambda = 433 nm is incident on a diffraction grating the first maximum after the center one is found to occur at an angle of theta_1 = 8.5 degrees. (a) Write an expression for the grating's line density, n, in terms of the given quantities. (b) Calculate this spacing in lines per centimeter? n = (c) Find the angle of the second order intensity maximum, theta_2, in degrees.

Part B & C

Answer 1

Salution PART1B) given t=433 nm = 433 8109 Oi=8.50 for m=) In cliffrection grating conelition for maxima is given as dsince = my . d= mi sino We know n=t o d = 1 ano na Bino mi for m=1 0,=8.50 n= sin 8.5° 1x 433x109 m m = 0.000341361 X 10' lines 10-2m In = 3413°61 lines CY) OR In=3414 lines,

- PART(C) for m=2 O₂ = 1 d SinO2 =21 sino, d= + sino, sino =21 sinon sino,= asino, 2= sinti (2 sin 8.5%) 0,= 17.19° Answer