91Ó°ÊÓ

The wavelength of the monochromatic light is $\mathrm{λ}=500\text{nm}$.

The width of slit is $\mathrm{a}=6\mathrm{μ³¾}$.

The distance of slit from screen is $D=3\text{m}$.

The distance from the central maxima is $y=15\text{cm}$.

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 for point P is given as:

$\tan \mathrm{θ}=\frac{y}{D}$

Substitute all the values in equation.

localid="1664272072563" $$\begin{gathered} \tan \mathrm{θ}=\frac{\left(15\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}{3\text{m}} \\ \tan \mathrm{θ}=0.05 \\ \mathrm{θ}=2.86° \end{gathered}$$

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

$\mathrm{Î\pm }=\left(\frac{\mathrm{Ï€}a}{\mathrm{λ}}\right)\sin \mathrm{θ}$

Substitute all the values in equation.

localid="1664272077973" $$\begin{gathered} \mathrm{Î\pm }=\left(\frac{\mathrm{Ï€}\left(6\mathrm{μ³¾}\right)\left(\frac{{10}^{-6}\mathrm{m}}{1\mathrm{μ³¾}}\right)}{\left(500\mathrm{nm}\right)\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)}\right)\left(\sin 2.86°\right) \\ \mathrm{Î\pm }=1.88\mathrm{rad} \end{gathered}$$

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

localid="1664272083528" $\frac{I_P}{I_m}={\left(\frac{\sin \mathrm{Î\pm }}{\mathrm{Î\pm }}\right)}^2$

Substitute all the values in equation.

localid="1664272088896" $$\begin{gathered} \frac{I_P}{I_m}={\left(\frac{\sin \left(\left(1.88\right)\left(\frac{180°}{\mathrm{π}}\right)\right)}{1.88}\right)}^2 \\ \frac{I_P}{I_m}=0.257 \end{gathered}$$

Therefore, the ratio of intensity of angular position P to the intensity at central maximum is $0.257$.

(b)

The distance between two minima is given as:

$w=\frac{\mathrm{λ}D}{a}$

Substitute all the values in equation.

$$\begin{gathered} w=\frac{\left(500\mathrm{nm}\right)\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)\left(3\mathrm{m}\right)}{\left(6\mathrm{μ³¾}\right)\left(\frac{{10}^{-6}\mathrm{m}}{1\mathrm{μ³¾}}\right)} \\ w=0.25\mathrm{m} \end{gathered}$$

Therefore, the distance between two minima is$0.25\text{m.}$