Question

A $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{mm}$wide slit is illuminated by light of wavelength $\mathbf{589}\mathbf{}\mathbf{nm}$. Consider a point P on a viewing screen on which the diffraction pattern of the slit is viewed; the point is at $\mathbf{30}\mathbf{°}$from the central axis of the slit. What is the phase difference between the Huygens wavelets arriving at point P from the top and midpoint of the slit? (Hint: See Eq. $\mathbf{36}\mathbf{-}\mathbf{4}$.)

Short answer

The path difference of diffraction is $85\pi \text{rad}$.

Step-by-step solution

Identification of given data

The wavelength of the light is $\lambda =589\text{nm}$.

The angle of diffraction minima from center of slit is $\theta =30°$.

The width of slit is $d=0.10\text{mm}$.

Concept used

The angle of the diffraction is defined as the angle of central axis from the diffraction minima of different orders.

Determination of path difference

The path difference of diffraction is given as:

$\Delta x=\frac{d}{2}\sin \theta$

Substitute all the values in equation.

$$\begin{gathered} \Delta x=\left(\frac{0.10\text{mm}}{2}\right)\left(\sin 30°\right) \\ \Delta x=0.025\text{mm} \end{gathered}$$

Determination of phase difference

The path difference of diffraction is given as:

$\Delta \varphi =\left(\frac{2\pi }{\lambda }\right)\left(\Delta x\right)$

Substitute all the values in equation.

$$\begin{gathered} \Delta \varphi =\left(\frac{2\pi }{\left(589\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)\left(0.025\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right) \\ \Delta \varphi =85\pi \text{rad} \end{gathered}$$

Therefore, the path difference of diffraction is $85\mathrm{\pi }\mathrm{rad}$.