# Question regarding physical optics - (thickness)

1. One of the two slits in a double-slit interference arrangement is closed with a thin strip of glass of refractive index 1.4 and the other slit is closed with a thin strip of transparent plastic with refractive index 1.7. On doing this, the centre of the interference pattern on the screen is seen to be displaced to the position of the fifth bright fringe (order 5) in the original pattern. If the wavelength of the monochromatic light used is 480 nm and if the strips have a thickness of t each, what is the value of t ?

I can't really understand with the question whehter I should use the Young Modulus double slit formula or the air wedge formula because Young Mudulus double slit does not involved refractive index ,n and in the air wedge,there is question on calculating the thickness.

So,I don't know what to do now because I get confuse with the question and have try so many times on calculation but still cannot get the answer.Help,I need somebody the help me figure out and explain to me what the question want.Thanks for anybody that spend sometime on this question.

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Please,I really need somebody help to explain to me what the question want so that I can have a better understanding of the question.Thanks for anybody that spend some time one this question.

OlderDan
Homework Helper
Sanosuke Sagara said:
1. One of the two slits in a double-slit interference arrangement is closed with a thin strip of glass of refractive index 1.4 and the other slit is closed with a thin strip of transparent plastic with refractive index 1.7. On doing this, the centre of the interference pattern on the screen is seen to be displaced to the position of the fifth bright fringe (order 5) in the original pattern. If the wavelength of the monochromatic light used is 480 nm and if the strips have a thickness of t each, what is the value of t ?

I can't really understand with the question whehter I should use the Young Modulus double slit formula or the air wedge formula because Young Mudulus double slit does not involved refractive index ,n and in the air wedge,there is question on calculating the thickness.

So,I don't know what to do now because I get confuse with the question and have try so many times on calculation but still cannot get the answer.Help,I need somebody the help me figure out and explain to me what the question want.Thanks for anybody that spend sometime on this question.
The derivation of the formula for double slit interference always assumes the light at the two slits is in phase. In general, you can have interference from two sources that are not in phase. The position of the interference maxima and minima depends on both the path difference from the sources, and on their phase difference. This fact is used by radio broadcasters to steer the direction of maximum intensity of signals. That is why you often see two transmission towers (sometimes four) close together.

In this situation, the film and the glass are going to result in a phase difference of the light exiting each slit. You need to figure out the amount of phase difference, and the thickness required to achive that phase difference based on the refractive indices of the two materials.

I think you shouldn't depend too much on the derived formula in the textbooks. Try to derive the formula base on the theory you learn.

I have drawn a diagram for you. I hope you can try again before reading my solution.

the first thing we can write down is the snell's law:
$$n_a \sin a' = \sin a$$ (1)

but the key idea is that the observed pattern on the screen is determined by the phase difference on the screen. In this particular case, the phase difference is contributed only by the optical path difference.

From the graph, you can easily convince yourself that the optical path difference of the two light rays is
$$\Delta x = \frac{n_a t}{\cos a'} - \frac{n_b t}{\cos b'}$$ (2)

The question is greatly simplified because we are considering the center of the screen, i.e. $$a = b = 0$$

such that $$\Delta x = \left( n_a - n_b \right) t$$,

and this is 5 lambda as required. All information is known and t can be determined.

p.s. with (1) and (2) you can work out the general case.

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OlderDan
Homework Helper
Kelvin said:
I think you shouldn't depend too much on the derived formula in the textbooks. Try to derive the formula base on the theory you learn.

I have drawn a diagram for you. I hope you can try again before reading my solution.

the first thing we can write down is the snell's law:
$$n_a \sin a' = \sin a$$ (1)

but the key idea is that the observed pattern on the screen is determined by the phase difference on the screen. In this particular case, the phase difference is contributed only by the optical path difference.

From the graph, you can easily convince yourself that the optical path difference of the two light rays is
$$\Delta x = \frac{n_a t}{\cos a'} - \frac{n_b t}{\cos b'}$$ (2)

The question is greatly simplified because we are considering the center of the screen, i.e. $$a = b = 0$$

such that $$\Delta x = \left( n_a - n_b \right) t$$,

and this is 5 lambda as required. All information is known and t can be determined.

p.s. with (1) and (2) you can work out the general case.
I don't think you need to bring Snell's law into this problem. The paths to any point on the screen must leave the slit/medium combination in essentially the same direction. You can just as well put the materials on the front side of the slits so that the propegation through the materials is perpendicular to the surfaces. Then the only effect is the phase shift in the materials due to a path length of exactly t before diffraction by the slits, with propegation from the slits through a single medium to arrive at the screen.

If the slits are very closely spaced, and the angles get large, then it will make a noticable difference if the materials come before or after the slit. Placing them in front of the slits removes the complication of the variable path length in the materials. If, as is usual, the angles are small, it doesn't make much difference, as your approximation of setting the cosines to 1 illustrates.

Thanks for both Kelvin and OlderDan help in this question.I really appreciate it and I will try to solve this question.