Q39P - University Physics with Modern Physics

Question

Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is 11.0 cm long and has a refractive index of 1.55. A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end, in a manner similar to that discussed in Example 35.4. When you view  the glass plates from above with reflected white light, you observe that, at 1.15 mm from the line where the sheets are in contact, the violet light of wavelength 400.0 nm is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. 

(a) How far from the line of contact will green light (of wavelength 550.0 nm) and orange light (of wavelength 600.0 nm) first be enhanced? 

(b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? 

(c) How thick is the metal foil holding the ends of the plates apart?

Step-by-Step Solution

Answer

a) 1.58 mm, 1.73 mm.

b) 3.45 mm, 4.75 mm, 5.18 mm.

c) $9.56\mu m$

Step 1: Given

$$\begin{gathered} L=11.0cm=11.0\times {10}^{-2}m \\ n_{plates}=1.55 \\ {\lambda }_{violet}=400nm=400\times {10}^{-9}m \\ {x}_{violet}=1.15mm=1.15\times {10}^{-3} m \\ {\lambda }_{green}=550nm=550\times {10}^{-9} m \\ {\lambda }_{orange}=600nm=600\times {10}^{-9} m \end{gathered}$$

Step 2: Solving part (a) of the problem.

Since the two plates are in contact from one side, which we chose to be the left one, and raised from the other side due to the thin metal foil, there is a very thin layer of air between the two plates. 

Now we are representing the reflected rays from the thin film of air with two rays, as you see below. 

The first ray experiences no phase change since the index of refraction of the air is less than the index of refraction of the plate. 

But the second ray experiences a phase change since the index of refraction of the second plate is greater than that of the air (The red circle indicates a phase change). 

So, we have two reflected rays with one phase change 

Hence, the thickness of the air film that makes the interference, between the two reflected rays, constructive is given by

$2t=\left(m+\frac{1}{2}\right){\lambda }_{film}$                                                    

In this case, the film is made of air. Hence,

$t=\frac{\left(m+{\displaystyle \frac{1}{2}}\right){\lambda }_{film}}{2}$                                                                                             (1)

             

 

Now can easily find the thickness of the air at the violet fringe. 

$t_{voilet }=\frac{\left(m+\frac{1}{2}\right){\lambda }_{voilet }}{2}$                                                       

The first bright fringe, which is at 115 mm from the left side (at the point in which the two plates are in contact), is the violet light (400 nm) and there are no bright fringes before this one. This means that m =0 0. 

Hence,

$\begin{array}{r}{\mathrm{t}}_{\mathrm{voilet} }=\frac{\left(0+\frac{1}{2}\right){\mathrm{\lambda }}_{\mathrm{voilet} }}{2}\\ {\mathrm{t}}_{\mathrm{voilet} }=\frac{{\mathrm{\lambda }}_{\mathrm{voilet} }}{4}\end{array}$                                                       

put the given,

$$\begin{gathered} {\mathrm{t}}_{voilet }=\frac{400\times {10}^{-9}}{4} \\ {\mathrm{t}}_{voilet }=100\times {10}^{-9} \end{gathered}$$                                                                                                 (2)

 

Since the thickness of the air in any position is not required, and we need to find the distance x of each fringe, we must find t in terms of x and .$\theta$ 

In the figure below, we represented only the first violet fringe only, but we will solve for a general formula after finding the value of $\theta$. From the right triangle, in the second figure below,

$\tan \mathrm{\theta }=\frac{{\mathrm{t}}_{voilet }}{x_{voilet }}$                                                                                                        (3)

$\theta ={\tan }^{-1}\left(\frac{{\mathrm{t}}_{voilet }}{x_{voilet}}\right)$                                       

Put the given and put from (2)

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{100\times {10}^{-9}}{1.15\times {10}^{-3}}\right) \\ \theta =\left(4.98\times {10}^{-3}\right) \end{gathered}$$                                                                                           (4)

The angle is too small.

                 

Now, in general, and from (3), we got

$t=\tan \theta x$                                                             

substitute into (1),

$\tan \theta x=\frac{\left(m+\frac{1}{2}\right){\lambda }_{air }}{2}$                                                       

solve for x;

$x=\frac{\left(m+\frac{1}{2}\right){\lambda }_{air }}{2\tan \theta }$                                                                                              (5)

Noting that the first bright fringe for each color means that m = 0. 

Hence,

$$\begin{gathered} x=\frac{\left(0+\frac{1}{2}\right){\lambda }_{cir }}{2\tan \theta } \\ x=\frac{{\lambda }_{\mathrm{air} }}{4\tan \theta } \end{gathered}$$                                                           

The boxed equation above is a general equation for this case. Now we will find each color.

For Green light:

$x_{green }=\frac{{\lambda }_{green }}{4\tan \theta }$                                                         

put the given and $\theta$ from (4)

$$\begin{gathered} x_{green }=\frac{550\times {10}^{-9}}{4\tan \left(4.98\times {10}^{-3}\right)} \\ {x}_{green }=1.58\times {10}^{-3}\mathrm{m}=1.58\mathrm{mm} \end{gathered}$$                                                     

For orange light:

$x_{orange }=\frac{{\lambda }_{orange }}{4\tan \theta }$                                                         

put the given and from (4)

$\begin{array}{c}{x}_{orange }=\frac{600\times {10}^{-9}}{4\tan \left(4.98\times {10}^{-3}\right)}\\ x_{\mathrm{orange} }=1.73\times {10}^{-3}\mathrm{m}=1.73\mathrm{mm}\end{array}$                                                     

Step 3: Solving part (b) of the problem.

The second enhancement for each color is for m =1.0 

Put into (5);

$\begin{array}{c}x=\frac{\left(1+\frac{1}{2}\right){\lambda }_{air}}{2\tan \theta }\\ x=\frac{1.5{\lambda }_{air }}{2\tan \theta }\end{array}$                                                         

For violet light:

${\mathrm{x}}_{\mathrm{violet} }=\frac{1.5{\mathrm{\lambda }}_{\mathrm{iolet} }}{2\tan \mathrm{\theta }}$                                                         

put the given and $\theta$ from (4)

$\begin{array}{c}{x}_{violet }=\frac{1.5\times 400\times {10}^{-9}}{4\tan \left(4.98\times {10}^{-3}\right)}\\ x_{violet }=3.45\times {10}^{-3}\mathrm{m}=3.45\mathrm{mm}\end{array}$                                                   

For Green light:

$x_{green }=\frac{1.5{\lambda }_{green }}{2\tan \theta }$                                                       

put the given and $\theta$ from (4)

$\begin{array}{c}{x}_{green }=\frac{1.5\times 550\times {10}^{-9}}{2\tan \left(4.98\times {10}^{-3}\right)}\\ x_{\mathrm{green} }=4.75\times {10}^{-3}\mathrm{m}=4.75\mathrm{mm}\end{array}$                                                   

For orange light:

$x_{orange }=\frac{1.5{\lambda }_{\mathrm{orange} }}{2\tan \theta }$                                                       

put the given and $\theta$ from (4)

                                              

Step 4: Solving part (c) of the problem.

We can find the thickness of the foil by the geometry of the right triangle in the figure above (since we know that angle and the length of the plate). 

Hence,

$\tan \theta =\frac{t_{foil}}{L}$                                                           

So, 

$t_{foil}=L\tan \theta$                                                           

put the given and $\theta$ from (4)

$$\begin{gathered} t_{foil}=11.0\times {10}^{-2}\times tan (4.98\times {10}^{-3}) \\ {t}_{foil}=9.56\times {10}^{-6} m=9.56\mu m \end{gathered}$$