Q15E - University Physics with Modern Physics

Question

A slit 0.240 mm wide is illuminated by parallel light rays of wavelength 540 nm. The diffraction pattern is observed on a screen that is 3.00 m from the slit. The intensity at the center of the central maximum 1u = 0_2 is 6.00 * 10-6 W>m2. 

(a) What is the distance on the screen from the center of the central maximum to the first minimum? 

(b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

Step-by-Step Solution

Answer

a) 6.75 mm

b) $I=2.43\times {10}^{-6} W/m^2$

Step 1: Given.

$a=0.240 mm=0.240\times {10}^{-3} m$, $\lambda =540 nm=540\times {10}^{-9} m$,$R=3.0m$,$I_0=6.0\times {10}^{-6} W/m^2$,m =1

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

We know that the distance y1 from the central maxima to the first minimum is given from

$\tan {\theta }_1=\frac{y_1}{R}$                                                           

So,

$y_1=Rtan {\theta }_1$                                                                                                          (1)

To find the distance ${\gamma }_1$ from the central maxima to the first minimum, we need first to find the angle ${\theta }_1$  

We know that the angle of the minimum fringes in single-slit experiments is given by

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

Noting that for the first minimum fringe m = 1, so 

$sin {\theta }_1=\frac{\lambda }{a}$                                                               

Solving for ${\theta }_1$

$$\begin{gathered} {\theta }_1={\sin }^{-1} \left(\frac{\lambda }{a}\right) \\ {y}_1=3.00\times \tan \left[{\sin }^{-1} \left(\frac{540\times {10}^{-9}}{0.240\times {10}^{-3}}\right)\right] \\ {y}_1=6.75\times {10}^{-3}\mathrm{m} \end{gathered}$$

Substitute into (1)

$y_1=Rtan\left[si{n}^{-1}\left(\frac{\lambda }{a}\right)\right]$

Substitute the given

$\begin{array}{c}{y}_1=3.00\times \tan \left[{\sin }^{-1} \left(\frac{540\times {10}^{-9}}{0.240\times {10}^{-3}}\right)\right]\\ y_1=6.75\times {10}^{-3}\mathrm{m}\end{array}$

Step 3: Calculate the intensity at a point on the screen midway between the center of the central maximum and the first minimum.

First of all, we need to find the point in which its position on the screen is midway between the center of the central maximum and the first minimum. 

This position is given by

$y=\frac{y_1}{2}$                                                                                                           (2)

We know that the intensity is given by 

$I=I_0{\left(\frac{\sin {\displaystyle }\frac{\beta }{2}}{\frac{\beta }{2}}\right)}^2$

We also know that 

$\beta =\frac{2\pi a\sin \theta }{\lambda }$

Substitute into the intensity formula above, noting that 2 cancels;

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

Now we need to find the angle $\theta$ of this new position y, which is given by

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

Substitute into (2)

$$\begin{gathered} \tan \theta =\frac{y_1}{2R} \\ \theta =ta{n}^{-1}\left(\frac{y_1}{2R}\right) \end{gathered}$$

Substitute the given and from the final result in part an above. 

$$\begin{gathered} \theta =ta{n}^{-1}\left(\frac{6.75\times {10}^{-3}}{2\times 3.0}\right) \\ \theta =0.{06446}^{\circ }=1.125\times {10}^{-3} rad \end{gathered}$$

Substitute  in rads and the given into (3). Do not forget to convert your calculator to RAD mode

$I=6.0\times {10}^{-6}\times {\left(\frac{\sin {\displaystyle }\left(\frac{\pi \times 0.240\times {10}^{-3}\sin \left(1.125\times {10}^{-3}\right)}{540\times {10}^{-9}}\right.}{\frac{\pi \times 0.240\times {10}^{-3}\sin \left(1.125\times {10}^{-3}\right)}{540\times {10}^{-9}}}\right)}^2$

\(I = 2.43 \times {10^{ - 6}}W/{m^2}\)