Question

Q: You pass 589.0-nm radiation and 589.6-nmradiation through a diffraction grating that has525.0 slits/mm.

Part A: In the diffraction pattern, what is the angular separation between them = 3 peak for the 589.0-nm radiation and the m = 3 peak for the589.6-nm radiation?

θ589.6nm−θ589.0nm= ?

Part B: Are there fourth-order fringes in the diffraction pattern?

yes?no?

Answer 1

A.)sin theta = m lambda/d

d is the slit separation

m is order

theta is angular seperation

lambda is wavelength

at m = 3 , lambda = 589.0-nm

d = 1/525 = 1.9047 * 10^-3 mm = 1.9047 * 10^-6 m

sin theta = 3 * 589 * 10 ^-9 m/ 1.9047 * 10^-6 m

theta = 68.07^o

at m = 3 , lambda = 589.6-nm

d = 1/525 = 1.9047 * 10^-3 mm = 1.9047 * 10^-6 m

sin theta = 3 * 589.6 * 10 ^-9 m/ 1.9047 * 10^-6 m

theta = 68.225 ^o

B.) check whether fourth-order fringes in the diffraction pattern is possible

at m = 4 , lambda = 589.6-nm

d = 1/525 = 1.9047 * 10^-3 mm = 1.9047 * 10^-6 m

sin theta = 3 * 589.6 * 10 ^-9 m/ 1.9047 * 10^-6 m

this is not possible since sin theta become 1.238 which is not possible

similarly for

at m = 4 , lambda = 589.0-nm

d = 1/525 = 1.9047 * 10^-3 mm = 1.9047 * 10^-6 m

sin theta = 4 * 589 * 10 ^-9 m/ 1.9047 * 10^-6 m

sin theta = 1.22 whisc is also not possible

fourth-order fringes in the diffraction pattern is not possible