In a wire YDSE experiment a thin film of thickness t1 and t2 are place...
Explanation:
In a YDSE experiment, interference fringes are obtained by splitting a light beam into two coherent beams using two slits. These beams then interfere with each other to produce a pattern of bright and dark fringes on a screen. When a thin film is placed in front of one or both of the slits, the interference pattern changes.
Experiment 1:
In the first experiment, a thin film of thickness t1 is placed in front of slit S1. The interference pattern is observed on a screen placed at a distance from the slits. The first minimum is observed at point a, which is symmetrical to S with respect to S1.
Experiment 2:
In the second experiment, a thin film of thickness t2 is placed in front of slit S2. The interference pattern is observed on a screen placed at a distance from the slits. The second maximum is observed at point a, which is symmetrical to S with respect to S2.
Calculation of x:
Let the wavelength of the light used be λ.
In the first experiment, the path difference between the two beams at point a is given by:
2t1 = (m + 1/2)λ, where m is an integer.
In the second experiment, the path difference between the two beams at point a is given by:
2t2 = (n + 1)λ, where n is an integer.
Since point a is symmetrical to S with respect to both S1 and S2, the path difference in both experiments should be equal. Therefore,
2t1 = 2t2
t2/t1 = 1/2
Given that t2/t1 = x/25, we have:
x/25 = 1/2
x = 12.5
Conclusion:
Therefore, if the ratio of the thickness of two thin films placed in front of slits S1 and S2 respectively is x/25 and the first minimum and second maximum are produced at a point in the first and second experiment respectively, then x = 12.5.