Q56CP - University Physics with Modern Physics

Question

Figure P35.56 shows an interferometer known as Fresnel’s biprism. The magnitude of the prism angle A is extremely small. (a) If S0 is a very narrow source slit, show that the separation of the two virtual coherent sources S1 and S2 is given by d = 2aA(n – 1), where n is the index of refraction of the material of the prism. (b) Calculate the spacing of the fringes of green light with wavelength 500 nm on a screen 2.00 m from the biprism. Take a = 0.200 m, A = 3.50 mrad, and n = 1.50

Step-by-Step Solution

Answer

1. We see $d=2aA(n-1)$
2. The distance is $x=1.57\times {10}^{-3}m$

Step 1: Important Concepts

Constructive interference is at $\mathit{n}\mathit{\lambda }$

and destructive interference is at $\frac{\left(2n+1\right)\mathbf{\lambda }}{\mathbf{2}}$.

Where n is an integer.

Step 2: Distance between the prisms

Let the distance between the first virtual image and the straight side be x

 

We know, Angle of incidence and angle of refraction on that straight side must be very small so a concentrated image can be formed

                                            $$\begin{gathered} \sin i\approx \tan i=\frac{w}{a} \\ \tan i\approx \tan r=\frac{w}{x} \end{gathered}$$         

                                                     

By using law of refraction

                                     $n_{i}sin i=n_{2}sin r$                  

We get

                                     $$\begin{gathered} n_{i}sin i=n_{2}sin r \\ \frac{w}{a}=n\frac{w}{x} \end{gathered}$$                  

Rearranging 

                                    $x=na$                                   

                                    

 

According to the equal the final virtual image is on the same horizontal position as the light source i.e.(A is small)

We also see that the two virtual images and the light source from a right angled triangle 

 

We also observe

                                   $$\begin{gathered} \tan A=\frac{{\displaystyle \frac{d}{2}}}{na-a})\approx A \\ d=2aA(n-1) \end{gathered}$$                  

                                                       

Step 3: Fringe distance

Find the distance between the 

                                                   

                                      $$\begin{gathered} d=2(0.2)(3.5\times {10}^{-3})(1.5-1) \\ d=7\times {10}^{-4} \end{gathered}$$                        

 

We know that 

                                      $\lambda =\frac{dx}{a+b}$                      

Solve for x

                           $x=\frac{500\times {10}^{-9}(0.2+2)}{7\times {10}^{-4}}=1.57\times {10}^{-3}m$                      

Hence, The distance is $x=1.57\times {10}^{-3} m$