Light with wavelength 400 nm passes through a diffraction grating with 10^-4 m b

Question

Light with wavelength 400 nm passes through a diffraction grating with 10^-4 m between adjacent cells. The intensity is observed on a large distant screen. (See the top of the page.) What is the total number of intensity maxima on the screen? 1 2 3 4 5 6 (See the top of the page) What angle theta locates the intensity maximum that is farthest from the horizontal line? 23.6 degree 53.1 degree 80.5 degree 90 degree As shown at the right, a semicircular crystal is surrounded by air. A light ray hits the center of the crystal's flat side as shown. The crystal's index of refraction is 1.414. When the light ray exits the crystal, what path doo it follow? path A (tangent to circle) path B (equidistant between path A and C) path C (radial direction) path D (equidistant between paths C and E) path E (tangent to circle)

Explanation / Answer

Q4. for maximum intensity, m*lambda=d*sin(theta)

where m=order of the intensity

lambda=wavelength=400 nm=4*10^(-7) ,m

d=slit width=10^(-6) m

then as maximum value of sin(theta) is 1,

maximum value of m=d/lambda=2.5

but as m is an integer, m=2 is the correct solution.

so we will have one central bright spot, and 2 bright spot on either side

so total maxima are 1+2+2=5

hence option E is correct.

Q5.for farthest maxima,

m=2

then 2*4*10^(-7)=10^(-6)*sin(theta)

==>sin(theta)=0.8

==>theta=53.1 degrees

hence option B is correct.

Q6.as incident angle at the semicircular surface is 0,

there will not be any refraction

hence path C will be taken

option C is correct.

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