Question

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.350 mm with the screen 39.0 cm away from the slit, when light of wavelength 570 nm is used.
(a) Find the slit width.
(b) Calculate the angle θ of the first diffraction minimum.

Answer 1

Answer: a) the width is 2.54 mm

b) the angle is 0.012°

Explanation:

we have the equation:

λ = zw/(L*m)

Where is the distance for the middle of the screen, L is the distance to the screen, w is the widht of the slit and λ is the wavelength.

now we can isolate z and get:

z = λL*m/w

and the distance between the 5 minimum and the first one is 0.035mm

then:

0.035mm = λL*5/w - λL*1/w = λL*4/w

now we can solve it for w.

0.35mm = 570nm*39.0cm*4/w

now, we have all diferent units, lets use nanometers.

1cm = 1x10^-2 m

1mm = 1x10^-3 m

1nm = 1x10^-9m

then: 1 cm = 1x10^7 nm

         1 mm = 1x10^6 nm

then we have:

0.35x10^6 nm = 570nm*39.0x10^7 nm*4/w

w = (570nm*39.0x10^7 nm*4)/0.35x10^6 nm =(570nm*39.0*10*4)/0.35

w = 2540571.4 nm

it is a bigg number, let's write it in milimeters:

w = (2540571.4/10^6) mm = 2.54 mm

the first minimum can be obtained by the equation w*sinθ = mλ by using m = 1

then we have:

2540571.4 nm*sinθ = 570 nm

sinθ = 570/2540571.4 = 0.000224

θ = asin(0.00022) = 0.012°