Question

A slit 0.240mm wide is illuminated by parallel light rays with a wavelength 600nm . The diffraction pattern is observed on a screen that is 3.50m from the slit. The intensity at the center of the central maximum (theta = 0) is 4.10�10?6 W/m^2. What is the distance on the screen from the center of the central maximum to the first minimum? What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?Expert Answer

Answer 1

To solve this exercise, it is necessary to apply the concepts on the principle of superposition, specifically on constructive interference,

Constructive interference (light spot) is defined by

$dsin\theta=m\lambda$

Where,

m = The integer m is called the interference order and is the number of wavelengths by which the two paths differ.

$\lambda = wavelenght$

d = Distance

For smaller angles $sin\theta = tan\theta$,

$d tan\theta = m \lambda$

From the trigonometric properties it is understood that so the is - in this case - the length measured vertically by reason of distance, that is

$d*\frac{y}{L} = m\lambda$

Re-arrange to find y,

$y = \frac{m\lambda L}{d}$

Replacing our values

$y = \frac{(1)(600*10^{-9}(3.5m)}{0.240*10^{-3}m}$

$y = 8.75*10^{-3}m$

$y = 8.75mm$

PART B) To calculate the intensity it is necessary to find the angle between the previously calculated height and distance in order to calculate the phase angle, in other words,

$tan\theta' = \frac{y}{L}$

$tan\theta' = \frac{1/2(8.75*10^{-3})}{3.5}$

$\tetha = 0.07161\°$

Therefore phase angle is

$\gamma = \frac{2\pi}{\lambda}sin\theta'$

$\gamma = \pi$

The intensity formula would then be given by,

$I = I_0 \frac{sin\frac{\gamma}{2}^2}{\frac{\gamma}{2}}$

$I = 4.1*10^{-6}(\frac{4}{\pi})$

$I = 1.66*10^{-6}W/m^2$