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A single slit of width 0.2 mm is illuminated by a mercury light of wavelength 365...

A single slit of width 0.2 mm is illuminated by a mercury light of wavelength 365 nm. Find the intensity at an 11° angle to the axis in terms of the intensity of the central maximum.

I tried the Im(sin (alpha)/alpha)^2 equation, where alpha is (pi(distance)(sin theta))/(wavelength) equation and its just not working.My calculator is in degrees. thanks for any help . My thought is that I don't know what Im is?  

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Homework Answers

Answer #1

answer) the formula for intensity is given by

I=I0(sin /)2.............1)

here Io=intensity of central maximum

and =pidsin/

so plugging the values to find

=pi*0.0002m*sin 11/365*10-9m=328.296 rad

now putting the values in eqn 1)( for this last part put your calculator in rad)

I=Io(sin328.296/328.296)=9.28*10-6Io

so the answer is I=9.28*10-6Io or I=9.3*10-6Io

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