Q54P - University Physics with Modern Physics

Question

A slit 0.360 mm wide is illuminated by parallel rays of light that have a wavelength of 540 nm. The diffraction pattern is observed on a screen that is 1.20 m from the slit. The intensity at the center of the central maximum $\left(\theta -0°\right)$ is ${\mathit{l}}_{\mathbf{0}}$ . (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to ${\mathit{l}}_{\mathbf{0}}$/2?

Step-by-Step Solution

Answer

(a) The distance on the screen from the center of the central maximum to the first minimum is 1.8mm.

(b) The distance is 0.8mm.

Step 1: Intensity I on the screen

The equation gives the intensity I on the screen, which is represented by the red line and is dependent on $\theta$

$I=I_0{\left(\frac{\sin {\displaystyle }(\pi a\sin \theta /\lambda )}{\pi a\sin \theta /\lambda }\right)}^2$

The equation gives the minima, which are the points of destructive interference.

$\sin \theta =\frac{m\lambda }{a}$

Step 2: The distance on the screen from the center of the central maximum to the first minimum

(a) The first minimum ${\theta }_{1m}$ is determined by the equation;

$\sin {\theta }_{1m}=\frac{\lambda }{a}=\frac{5.4\times {10}^{-7}}{3.6\times {10}^{-4}}=\frac{3}{2000}$

similarly, the angle is determined by the equation in the figure.

$\tan {\theta }_{1m}=\frac{b}{x}$

As approximating the value;

$$\begin{gathered} b=x\tan {\theta }_{1m} \\ =x\sin {\theta }_{1m} \\ =1.2\times \frac{3}{2000}\mathrm{m} \\ =1.8\mathrm{mm} \end{gathered}$$

Hence, the distance on the screen from the center of the central maximum to the first minimum is 1.8mm

Step 3: The distance on the screen from the center of the central maximum to the point where the intensity has fallen to l 0 /2

(b) Using the intensity equation for the angle $\theta$ at which $l=l_0/2$

$\frac{1}{2}={\left(\frac{\sin {\displaystyle }\gamma }{\gamma }\right)}^2\Rightarrow \gamma =\sqrt{2}\sin \gamma$

Using the function f;

$f\left(\gamma \right)=\sqrt{2}\sin \gamma$

From the graph, there are two fixed points, thus the sequence is;

$$\begin{gathered} {\gamma }_2=f\left({\gamma }_1\right) \\ ... \\ {\gamma }_n=f\left({\gamma }_{n-1}\right) \end{gathered}$$

By following;

$$\begin{gathered} lim y_n=y_0 \\ n\to \infty \end{gathered}$$ 

So, the sufficient condition is;

$\left|f\text{'}\left({\gamma }_0\right)\right|<1$

 

As in the figure slope of f is smaller than that of the slope of identity (1), 

So, the sequences are;

$$\begin{gathered} {\gamma }_2=\sqrt{2} \\ {\gamma }_3=1.397 \\ {\gamma }_4=1.393 \\ {\gamma }_5=1.392 \\ {\gamma }_0=\underset{n\to \mathrm{\infty }}{lim} {\gamma }_n \\ =1.391557 \end{gathered}$$

From the definition;

${\gamma }_0=\frac{\pi a}{\lambda }\sin {\theta }_0\Rightarrow \sin {\theta }_0=\frac{\lambda {\gamma }_0}{\pi a}$

So, the distance is;

$$\begin{gathered} b_{\frac{1}{2}}=x\tan {\theta }_0 \\ =x\frac{{\gamma }_0}{\pi }\frac{\lambda }{a} \\ =1.2\frac{1.391557}{\pi }\frac{3}{2000}\mathrm{m} \\ \approx 0.8\mathrm{mm} \end{gathered}$$

Hence, the distance is 0.8mm.