Question

Light with a wavelength of 587.5 nm illuminates a single slit 0.750 mm in width.
(a.) At what distance from the slit should a screen be located if the first minimum in the diffraction pattern is to be 0.850 mm from the center of the screen?
b.) What is the width of the central maximum?
(c.) Sketch the intensity distribution for the diffraction pattern observed on the viewing screen.

Answer 1

Answer:

a) The screen should be located at 1.08 meters

b) The width of the central maximum is 1.7 mm

c) See figure below.

Explanation:

a) This is a single slit diffraction problem, the equation that describes this kind of phenomenon is:

$a\sin\theta=m\lambda$ (1)

Because we’re interested in a minimum near the center of the screen, we can use the approximation $\sin\theta\approx\tan\theta=\frac{y}{x}$

So equation (1) is now:

$a\frac{y}{x}=m\lambda$ (2)

Solving (2) for x:

$x=\frac{ay}{m\lambda}=\frac{(0.75\times10^{-3})(0.85\times10^{-3})}{1(587.5\times10^{-9})}\approx1.08m$

b) As you can see on the figure below a maximum is approximately between the two adjacent minimums, because the diffraction pattern is approximately symmetric respect the center of the screen the width of the central maximum is 2*0.850mm = 1.7 mm.