Monochromatic light with wavelength $\mathbf{538}\mathbf{}\mathbf{nm}$is incident on a slit with width role="math" localid="1663172390189" $\mathbf{0}\mathbf{.}\mathbf{025}\mathbf{}\mathbf{mm}$. The distance from the slit to a screen is $\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{m}$. Consider a point on the screen $\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{cm}$from the central maximum. Calculate (a) $\mathit{\theta }$ for that point, (b) $\mathit{\alpha }$, and (c) the ratio of the intensity at that point to the intensity at the central maximum.

The wavelength of the monochromatic light is $\lambda =538\text{nm}$.

The width of slit is $a=0.025\text{mm}$.

The distance of slit from screen is $D=3.5\text{m}$.

The distance from the central maxima is $y=1.1\text{cm}$.

The angle of the diffraction is defined as the angle of central axis from the diffraction minima of different orders. The intensity of light in diffraction by double slits includes interference factor and diffraction factor.

(a)

The angle of diffraction is given as:

$\tan \theta =\frac{y}{D}$

Substitute all the values in equation.

$$\begin{gathered} \tan \theta =\frac{\left(1.1\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}{3.5\text{m}} \\ \tan \theta =0.003143 \\ \theta =0.18° \end{gathered}$$

Therefore, the angle of diffraction is $0.18°$.

(b)

The angle for different angular position on screen is given as:

$\alpha =\left(\frac{\pi a}{\lambda }\right)\sin \theta$

Here, $\theta$ is diffraction angle.

Substitute all the values in equation.

$$\begin{gathered} \alpha =\left(\frac{\pi \left(0.025\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(538\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)\left(\sin 0.18°\right) \\ \alpha =0.46\text{rad} \end{gathered}$$

Therefore, the angle for different angular position on screen is $0.46\text{rad}$.

(c)

The ratio of intensity of angular position to the intensity at central maximum is given as:

$\frac{I}{I_c}={\left(\frac{\sin \alpha }{\alpha }\right)}^2$

Substitute all the values in equation.

$$\begin{gathered} \frac{I}{I_c}={\left(\frac{\sin \left(\left(0.45\right)\left(\frac{180°}{\pi }\right)\right)}{0.45}\right)}^2 \\ \frac{I}{I_c}=0.93 \end{gathered}$$

Therefore, the ratio of intensity of angular position to the intensity at central maximum is $0.93$.