Question

In the single-slit diffraction experiment of $\mathit{F}\mathit{i}\mathit{g}\mathbf{.}\mathbf{36}\mathbf{-}\mathbf{4}$,let the wavelength of the light be $\mathbf{500}\mathbf{}\mathbf{nm}$, the slit width be localid="1664272054434" $\mathbf{6}\mathbf{\mu m}$, and the viewing screen be at distance localid="1664272062951" $\mathbf{D}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$. Let y axis extend upward along the viewing screen, with its origin at the center of the diffraction pattern. Also let ${\mathit{I}}_{\mathbf{p}}$represent the intensity of the diffracted light at point P at $\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$. (a) What is the ratio of ${\mathit{I}}_{\mathbf{p}}$to the intensity ${\mathit{I}}_{\mathbf{m}}$ at the center of the pattern? (b) Determine where point P is in the diffraction pattern by giving the maximum and minimum between which it lies, or the two minima between which it lies.

Short answer

(a) The ratio of intensity of angular position P to the intensity at central maximum is $0.257$.

(b) The distance between two minima is $0.25\text{m}$.

Step-by-step solution

Identification of given data

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

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

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

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

Concept used

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.

Determination of ratio of intensity at point P to the intensity at central maximum

(a)

The angle of diffraction for point P is given as:

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

Substitute all the values in equation.

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

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

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

Substitute all the values in equation.

localid="1664272077973" $$\begin{gathered} \alpha =\left(\frac{\mathrm{\pi }\left(6\mathrm{\mu m}\right)\left(\frac{{10}^{-6}\mathrm{m}}{1\mathrm{\mu m}}\right)}{\left(500\mathrm{nm}\right)\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)}\right)\left(\sin 2.86°\right) \\ \alpha =1.88\mathrm{rad} \end{gathered}$$

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

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

Substitute all the values in equation.

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

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

Determination of distance between two minima

(b)

The distance between two minima is given as:

$w=\frac{\lambda D}{a}$

Substitute all the values in equation.

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

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