Q14E - University Physics with Modern Physics

Question

Monochromatic light of wavelength λ = 620 nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit. In terms of the intensity I0 at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: 

(a) 1.00 mm; 

(b) 3.00 mm; 

(c) 5.00 mm?

Step-by-Step Solution

Answer

1. $0.822I_0$
2. $0.113I_0$
   

      c. $0.0253I_0$

Step 1 Given .

wavelength A= 620 nm, distance from the screen D = 3 m, and width of the slit are d=0.450 mm. The intensity at the central maximum is $l_0$

Step 2: Concept .

Let x be the distance of a point with the intensity I from the central maximum. Then we can represent this intensity as.

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

Now the relationship between x and D can be expressed in terms of $\sin \theta$ as 

                                                        $\sin \theta =\frac{x}{D}$      

For a very small angle, we can write $\theta$ as $\theta$. So the above equation becomes 

                                                      $\theta =\frac{x}{D}$             

Step 3: calculate the intensity of the light.

When the distance from the central maximum is x =1 mm, by substituting the value of  in terms of intensity, we get

                          $$\begin{gathered} I=I_0{\left[\frac{(\sin {\displaystyle \frac{\left(\pi d\right)\sin \left(\theta \right)}{\lambda }}}{\frac{\left(\pi d\right)\sin \left(\theta \right)}{\lambda }}\right]}^2 \\ I=I_0{\left[\frac{(\sin {\displaystyle \frac{(\pi 0.450\times {10}^{-3}) (1\times {10}^{-3})}{\left(620\times {10}^{-9}\right)\left(3\right)}}}{\frac{(\pi 0.450\times {10}^{-3}) (1\times {10}^{-3})}{\left(620\times {10}^{-9}\right)\left(3\right)}}\right]}^2 \\ =I_0{\left[\frac{\sin \left(0.760\right)}{0.760}\right]}^2 \\ =0.8221l_0 \end{gathered}$$

When the distance from the central maximum is x = 3 mm, by substituting this value in the above intensity equation, we get

                         $$\begin{gathered} I=I_0{\left[\frac{(\sin {\displaystyle \frac{(\pi 0.450\times {10}^{-3}) (1\times {10}^{-3})}{\left(620\times {10}^{-9}\right)\left(3\right)}}}{\frac{(\pi 0.450\times {10}^{-3}) (1\times {10}^{-3})}{\left(620\times {10}^{-9}\right)\left(3\right)}}\right]}^2 \\ =I_0{\left[\frac{\sin \left(2.27\right)}{2.27}\right]}^2 \\ =0.1131l_0 \end{gathered}$$

When the distance from the central maximum is x=5 mm, by substituting this value in the above intensity equation, we get

                       $$\begin{gathered} I=I_0{\left[\frac{(\sin {\displaystyle \frac{(\pi 0.450\times {10}^{-3}) (1\times {10}^{-3})}{\left(620\times {10}^{-9}\right)\left(3\right)}}}{\frac{(\pi 0.450\times {10}^{-3}) (1\times {10}^{-3})}{\left(620\times {10}^{-9}\right)\left(3\right)}}\right]}^2 \\ =I_0{\left[\frac{\sin \left(3.79\right)}{3.79}\right]}^2 \\ =0.0253l_0 \end{gathered}$$