Q13E - University Physics with Modern Physics

Question

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. 

(a) If the first diffraction minima are at _90.0_, so the central maximum completely fills the screen, what is the width of the slit? 

(b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at u = 45.0_ to the intensity at u = 0?

Step-by-Step Solution

Answer

a) 580 nm

b) 0.128

Step 1: Given.

$\lambda =580 nm=580\times {10}^{-9} m.{\theta }_1=\pm {90}^{\circ },\theta =45.0^{\circ }$

Step 2: Ray diagram.

Step 3: Calculate the width of the slit.

We know that the angle of the minimum fringe in the single-slit experiment is given by 

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

And in the case of the first minimum fringe, m =1;

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

Solving for a,

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

Substitute the given,

$$\begin{gathered} a=\frac{580}{sin {90}^{\circ }} \\ a=580nm \end{gathered}$$                                                             

Step 4: Calculate the ratio of the intensity .

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;

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

To find the ratio of the intensity at $\theta =45.0^{\circ }$ to the ratio of the intensity at $\theta =0^{\circ }$, we need to find $\frac{I}{I_0}$ since the intensity at an angle is the maximum (To). Hence,

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

Noting that a = $\lambda$ as we found in Step 3 above. So,

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

Substitute the given and note that $45.0°=\frac{\mathrm{\pi }}{4}$  

Remember to convert your calculator to RAD mode.

$$\begin{gathered} \frac{I}{I_0}={\left(\frac{\sin \left(\pi \sin \frac{\pi }{4}\right)}{\pi a\sin \frac{\pi }{4}}\right)}^2 \\ \frac{I}{I_0}=0.128 \end{gathered}$$