Consider the intensity distribution due to a 4-slit grating (Hint: we derived th
ID: 1550361 • Letter: C
Question
Consider the intensity distribution due to a 4-slit grating (Hint: we derived the result for an N-slit case in the class) with the spacing between adjacent. The spacing between two adjacent slits is 2.0 um and light of wavelength 500-nm is used in the experiment. Find the number, location and peak intensity of the principal maxima that may be observed on a screen 2.5 m away if the intensity contributed by each slit is I0. How many secondary intensity maxima are there between two adjacent principal maxima? Find the location and peak intensities of all the secondary maxima between the central and the first principal maxima on one side. How many interference minima are there between the above two principal maxima and where are they located?
Explanation / Answer
Find the number, location and peak intensity of the principal maxima that may be observed on a screen 2.5 m away if the intensity contributed by each slit is I0.(ten i think)
Answer: Central peak of intensity at =0 in diffraction grating with N slits is N2 times the central intensity from individual slit. Thus. 4^2*10=16*10=160 of principal maximum.or the while the height of each maxima is proportional to N2 .
For many slits, maxima are still at =sin(theta)=m*wavelength/spacing=1*500*10^(-9)/2*10^(-6)=0.2500
(first principal maxima) thus theta = 14.3239, Therefore, the smaller spacing the larger the angle . or from centre =D*tan(theta)=2.5*tan(14.32)= 0.6382 m
How many secondary intensity maxima are there between two adjacent principal maxima?
Answer : Between two principal maxima there are N-1 zeros (minima) and N-2 secondary maxima thus 4-2=2
Find the location and peak intensities of all the secondary maxima between the central and the first principal maxima on one side.
Between two principal maxima there are N-1 zeros (minima)