How Thin Should the Glass Be to Shift the Interference Maxima?

[Aphrael]

Homework Statement

A pair of slits separated by d = 1.30 mm is illuminated with light of l = 610 nm wavelength and falls on a screen L = 2.00 m away. A piece of glass with index of refraction n = 2.2 is placed at one slit. If the maxima shift is Dx/2, and falls on a minimum, what was the minimum glass thickness (in meters)?


2. Homework Equations / Attempt at solution

What is the significance of the Dx/2 shift maxima? Do I use the equation t =(m + 1/2)(wavelength) / 2n?

I know that wavelength is 610 x 10^-9 m; n=2.2, but I'm not sure whether I'm trying to use the wrong equation because of the extra information.

I'm also not sure whether I'm supposed to calculate a new wavelength of the light in the glass with n=2.2. If so, is it true that the wavelength(glass) = wavelength(air) / n(glass), therefore meaning that the new wavelength is (610nm)/(2.2) = 277.273 nm ?

 

[anathema]

It seems like you are on the right track with your equations and understanding of the problem. The maxima shift of Dx/2 is significant because it tells us that the shift is half of the distance between the slits, which is d = 1.30 mm. This means that the shift is equal to 0.65 mm.

In terms of the equation, you are correct in using t = (m + 1/2)(wavelength) / 2n. However, since the maxima shift is given in terms of distance, we can rewrite the equation as Dx/2 = (m + 1/2)(wavelength) / 2n. By plugging in the given values, we can solve for the minimum thickness of the glass:

0.65 mm = (m + 1/2)(610 x 10^-9 m) / (2 x 2.2)

Solving for m, we get m = 0.5. This means that the minimum thickness of the glass is when the shift of the maxima is half of the distance between the slits, which is 0.65 mm.

In regards to the wavelength in the glass, you are correct in using the equation wavelength(glass) = wavelength(air) / n(glass). However, since we are looking for the minimum thickness of the glass, we can assume that the light is traveling through the glass at an angle close to 90 degrees, which means the index of refraction will not affect the wavelength. Therefore, the new wavelength would still be 610 nm.

I hope this helps clarify your understanding of the problem. Keep up the good work in your scientific studies!

 

[Moetasim]

I would first clarify the question with the person who assigned the homework. The given information is not enough to solve the problem as it is missing the details of the interference pattern (e.g. whether it is a double-slit or single-slit experiment), the angle of incidence, and the specific location of the minimum on the screen. Additionally, the given equation for the minimum thickness of the glass is not accurate for thin film interference, as it is for the case of a single-slit experiment.

To answer your questions, the shift in maxima may be referring to the displacement of the interference pattern due to the presence of the glass, which can be calculated using the given wavelength and index of refraction. However, without knowing the specifics of the experiment, it is difficult to determine what the Dx/2 shift represents.

Regarding the calculation of the new wavelength in the glass, you are correct that the wavelength in the glass will be shorter due to the higher index of refraction. However, again, without knowing the specifics of the experiment, it is difficult to determine if this calculation is necessary or relevant to solving the problem.

In summary, more information is needed to accurately solve the problem and I would suggest seeking clarification from the person who assigned the homework. As a scientist, it is important to have all the necessary information and details before attempting to solve a problem.