Q26E - University Physics with Modern Physics

Question

A plate of glass 9.00 cm long is placed in contact with a second plate and is held at a small angle with it by a metal strip 0.0800 mm thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 nm. How many interference fringes are observed per centimeter in the reflected light?

Step-by-Step Solution

Answer

27.1 interference fringes are observed per centimeter in the reflected light.

Step 1: Given .

$L=9.00cm=9.00\times {10}^{-2}m$,$t=0.800mm=0.800\times {10}^{-3}m,\lambda =656nm=656\times {10}^{-9}m$

Step 2: Ray diagram.

                               

Step 3: Concept.

When the incident light falls on the lower surface of the upper piece of glass, it reflects with no phase change. 

But when the light falls on the upper surface of the lower piece of glass it reflects without a phase change (since the index of refraction of the glass is greater than that of the air). 

So, the reflected two rays from this system are having only one phase change, as you see in the second figure above ray 2. 

The red circle indicates a phase change. 

Since the left sides of the two glass plates are in touch, so the first fringe from the left side is a dark fringe. 

Now we need to find the distance between the two dark fringes, but first, we need to find the angle  between the two plates. 

From the geometry of the last figure above, the Right- triangle, we find that 

                          $\sin \theta =\frac{t}{L}$                                                                                     (1) 

Since we have one phase-change, and so the thickness that gives a dark fringe is given by

                          $2t=m\lambda$                                                

Hence,

                        $t=\frac{m\lambda }{2}$                                                                                           (2)

Now we need to know whether the other end that is laying on the metal strip is giving a bright or a dark fringe.

So we will solve (2) for m, if the answer was an integer number of m, so the last fringe is a dark one

                        $\mathrm{m}=\frac{2\mathrm{t}}{\mathrm{\lambda }}$                                       

Step 4: Calculation.

Substitute the given;

                                     $$\begin{gathered} m=\frac{2\times 0.0800\times {10}^{-3}m}{656\times {10}^{-9}m} \\ m=243.9\approx 244 \end{gathered}$$                

This means that the last fringe at the right end of the upper glass plate is a dark one. Now we need to find how many fringes in 10 cm.

    

N= Total number of fringes through the whole L plate / L

                                      $$\begin{gathered} N=\frac{244fringe}{9.0cm} \\ N=27.1fringe/cm \end{gathered}$$