Q2E - University Physics with Modern Physics

Question

Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit covering a lens with a focal length of 60.0 cm. In the focal plane of the lens, the distance from the central maximum to the first minimum is 8.65 mm. What is the width of the slit?

Step-by-Step Solution

Answer

the slit is 37.9 µm wide.

Step 1: Given Data

It is given that,

• Wavelength of light source, $\lambda =546 nm=546\times {10}^{-9}\hspace{0.33em}m$  
• Distance between screen and slit, R = 60 cm = 0.6 m  
• Distance of first minima from central maxima, $y=8.65mm=8.65\times {10}^{-3}\hspace{0.33em}m$

Step 2: Concept

The turning around of light rays from the edges of the object and entering the shadow region creating an interference pattern is known as diffraction.

Step 3: Width of Slit

By using the equation:

                                           $$\begin{gathered} \sin \theta =\frac{m\lambda }{a} \\ \lambda =\frac{a \sin \theta }{m} \end{gathered}$$
 

Here,  a  is the slit width, $\lambda$  is the wavelength of light used and  $\theta$ is the diffraction angle.

For first minima, m = 1  

 

                                            $\lambda =a \sin \theta$

 

To obtain the value of θ

 

 
           $$\begin{gathered} tan \theta =\frac{y}{R} \\ \theta =ta{n}^{-1}\left(\frac{y}{R}\right) \\ \theta =ta{n}^{-1} \left(\frac{8.65\times {10}^{-3}\hspace{0.33em}m}{0.6 m}\right) \\ \theta =0.826° \end{gathered}$$
 
 

 

Therefore, the wavelength of source is given by

 

 $$\begin{gathered} a=\frac{\lambda }{\sin \theta } \\ =\frac{546\times {10}^{-9 }m}{\sin \left(0.826°\right)} \\ =3.79\times {10}^{-5} m \\ =3.79 \mu m \end{gathered}$$

 

Therefore, the width of the slit is $3.79 \mu m$.