Question

(8.0.10) In class we showed that intensity distribution for single slit diffraction is given by I(θ) = I(0) [sin( β/ 2)/( β/2)]. ( β = ka sin(theta) , "a" is the slit width and k = 2π/λ ). (a) Show that the maxima in the diffraction pattern occur at θ for which tan( β/ 2) = β/2. (b) An obvious solution to this equation is β = 0, which corresponds to the central maximum. By sketching the functions tan(β/2) and β/2 as functions of β/2 on the plot, roughly estimate two values of β/2 other than zero which satisfy this equation. (c) Either by trial and error, numerical analysis or by asking your fancy calculators, find the smallest two positive values of β/2 which satisfy the condition derived in (a) to three significant figures.

(8.0.20) Light with wavelength λ = 600 nm diffracts through a slit with width 0.2 mm. (a) Using your results from the problem 8.0.10, find the values of θ for which the five maxima closest to the center occur. (You may include positive, negative, and zero values for theta.) (b) If the electric field at the center of the diffraction pattern oscillates with amplitude 60V/m, find the intensity at each of the maxima you identified in part (a).

You have to answer 8.0.10 and 8.0.10 because they are related!

Answer 1

Solution:

given that

In class we showed that intensity distribution for single slit diffraction is given by I(θ) = I(0) [sin( β/ 2)/( β/2)]. ( β = ka sin(theta) , "a" is the slit width and k = 2π/λ )..