Question

Light of wavelength 653.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 70.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.43 cm. What is the width of the slit?

Answer 1

.- The single slit diffraction pattern is given by

I (angle) = Io sinc^2 (d sin(angle) / lambda)

where the angle is given by

sin(angle) = x / L

where L is the distance from the slit to the observation screen and x is a position in the observation screen referenced at the maximum intensity position, that is I(x=0) = Io. Thus, the intensity distribution is given by

I (x) = Io sinc^2 (d x/L / lambda)

The first zero of the sinc function intensity distribution occurs at

d xo/L / lambda = 1

thus, xo is

d = lambda L / xo = lambda L / D/2

where D is the width of the central maximum. Then

d =0.000000653 * 0.705 / 0.0143 / 2 = 0.00064386 m