Problem 6: When light of a wavelength lambda = 402 nm is incident on a diffracti

Question

Problem 6: When light of a wavelength lambda = 402 nm is incident on a diffraction grating the first maximum after the center one is found to occur at an angle of theta1 = 3 degrees. Randomized Variables lambda = 402 nm Theta 1 = 3 degrees Part (a) Write an expression for the grating's line density, n, in terms of the given quantities. Expression : Select from the variables below to write your expression. Note that all variables may not be required. cos(a), cos(phi), cos(theta), cos(theta 1), sin(a), sin(phi), sin(theta), sin(theta 1), a, lambda, theta, d, g. m, t Part (b) Calculate this spacing in lines per centimeter? Numeric : A numeric value is expected and not an expression. Part (c) Find the angle of the second order intensity maximum, theta 2, in degrees. Numeric : A numeric value is expected and not an expression. Theta 2=

Explanation / Answer

In interfreence or diffraction pattern,

the needed equation is Y = mLR/d---------------1

and d sin theta = mL--------------------2

where L = wavelgnth,

m = order = 1,2,3,4, ......... for brigth bands

m = 1.5, 2.5, 3.5, 4.5, ......for dark bands

R is the distance from slit to screen,

Y = disatnce from central spot to nth order fringe

so

a. d sin theta    = mL

so

so d = 1 * 402 e-9/(sin 3)

d = 7.681 um
-----------------

b. No. of lines N = 1/d = 1/7.681e-

N = 130191 lines per m
-----------------------

c.

d sin theta = mL

sin theta = 2* 402 e-9/7.681 e-6

sin theta = 0.1046

theta =   6 deg

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