Interference pattern formed by an air wedge.

[semc]

An air wedge is formed between two glass plates separated at one edge by a very fine wire. When the wedge is illuminated from above by a 600nm light and viewed from above, 30 dark fringes are observed. Calculate the radius of the wire.

Am I correct to consider only the bottom surface of the top plate and the top surface from the bottom plate for the interference pattern? By considering that, I got $$2t=\frac{m\lambda}{2}$$. But what do I do after that?

 

[hermy]

hi semc

the wedge is illuminated from above. so, traveling through glass walls has no effect on light waves. no additional path difference is caused. it is perfectly alright to consider only the middle surfaces.

how did you arrive at that conclusion? what is t, m?

 

[semc]

It is supposed to be 2t=m.lambda instead of what I wrote.(For some weird reason I can't use the latex ) t would be the thickness of the air wedge, the space between the plates, and m is the order of interference where m takes on positive integer values. Let the top plate be the 1st plate and the bottom plate be the 2nd plate. The refracted ray from the bottom surface of the 1st plate undergoes no phase while the reflected ray from the top surface of the 2nd plate undergoes 180^o phase change. So the path difference in the air wedge should be integer values of the wavelength? But I have no idea how to go on after this. By the way why do we not consider the rays coming from the top surface of the 1st plate and the rays from the bottom surface of the 2nd plate?

 

[hermy]

We are viewing the whole experiment from the top. So we consider the rays coming from the top surface of the 1st plate.

One ray gets reflected from air (so no phase change) and the other travels through the glass, gets reflected from the 2nd plate (i.e a phase change of 180 = path change of $$\lambda$$, and again travels up through the 1st plate.

so the net path difference = 2t + $$\lambda$$ = m $$\lambda$$

(instead of just 2t, as you had earlier written)i guess, if light is incident close to the wire, then the diameter of wire would be equal to the thickness of wedge.
now you would know what to do next...