JEE Exam  >  JEE Questions  >  In a wire YDSE experiment a thin film of thic... Start Learning for Free
In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x?
Most Upvoted Answer
In a wire YDSE experiment a thin film of thickness t1 and t2 are place...
Community Answer
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.
Explore Courses for JEE exam

Similar JEE Doubts

In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x?
Question Description
In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x?.
Solutions for In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x? defined & explained in the simplest way possible. Besides giving the explanation of In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x?, a detailed solution for In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x? has been provided alongside types of In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x? theory, EduRev gives you an ample number of questions to practice In a wire YDSE experiment a thin film of thickness t1 and t2 are placed in front of slits S1 and S2 shown in the figure 1 and figure 2. It is observed that first minima and second maximum are produced at a point in first and second experiment respectively. Point a and S are symmetrical with respect to S1 and S2. Both the films have same refractive index if t2/t1 = x/25, then calculate x? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev