In Youngs double slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000 coming from the coherent sources S1 and S2. At certain point P on the screen third dark fringe is formed. Then the path difference S1 p − S2pin microns isa)0.75b)1.5c)3.0d)4.5Correct answer is option 'B'. Can you explain this answer?

Question

In Youngs double slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000  coming from the coherent sources S1 and S2. At certain point P on the screen third dark fringe is formed. Then the path difference S1 p &minus; S2pin microns isa)0.75b)1.5c)3.0d)4.5Correct answer is option 'B'. Can you explain this answer?

Answer

And S2 p is equal to 3 times the wavelength of the light.

We know that the distance between two consecutive bright fringes is given by:

d = λD/d

Where λ is the wavelength of light, D is the distance between the double-slit and the screen, and d is the distance between the two slits.

Similarly, the distance between two consecutive dark fringes is also given by:

d = λD/2d

Here, we are interested in the third dark fringe, which means that there are 3 bright fringes between the two slits and the point P. Therefore, the path difference between the two slits and the point P must be equal to 3λ.

Let's assume that the distance between the two slits is d. Then, the distance between the first bright fringe and the central fringe is given by:

y = λD/d

The distance between the third bright fringe and the central fringe is:

y' = 3λD/d

The path difference between the two slits and the point P is:

S1P - S2P = y' - y = 2λD/d

But we know that S1P - S2P = 3λ

Therefore, we can write:

2λD/d = 3λ

d = 2D/3

So the path difference between S1 and P or S2 and P is:

S1P = S2P = d/2 = D/3

Therefore, the path difference S1P and S2P is equal to D/3.

Answer

The path difference between S1 and S2 to point P on the screen where the third dark fringe is formed is:

Δx = (3/2)λ

where λ is the wavelength of the light.

Therefore, the path difference between S1 and P is half of that:

S1P = (3/4)λ

So, the path difference between S1 and P is (3/4) times the wavelength of the light.

Answer

The path difference between S1 and S2 at the third dark fringe on the screen can be calculated using the formula:

path difference = (m + 1/2)λ

where m is the order of the dark fringe (m = 0 for the central bright fringe, m = 1 for the first dark fringe, m = 2 for the second bright fringe, and so on), λ is the wavelength of the light, and 1/2 is added for the dark fringes.

In this case, we are given that the third dark fringe is formed, so m = 2. The wavelength of the light is 6000 angstroms, which is equal to 600 nm or 6 x 10^-7 m. Substituting these values into the formula, we get:

path difference = (2 + 1/2)(6 x 10^-7 m) = 1.5 x 10^-6 m

Therefore, the path difference between S1 and P at the third dark fringe on the screen is 1.5 x 10^-6 m.

Answer

And S2 p is given by:

path difference = (3/2)λ

Here's how to derive this equation:

In the double slit experiment, the light from S1 and S2 interferes on the screen to form a pattern of bright and dark fringes. The position of the fringes depends on the path difference between the two sources.

At the central maximum, the path difference is zero, and the bright fringe is formed. At the first dark fringe, the path difference is λ/2, at the second dark fringe, the path difference is λ, and so on.

To find the path difference at the third dark fringe, we can use the equation:

path difference = mλ/2

where m is the order of the fringe (m = 1 for the first dark fringe, m = 2 for the second dark fringe, etc.)

For the third dark fringe, m = 3, so we have:

path difference = (3/2)λ

Therefore, the path difference between S1 and P and between S2 and P at the third dark fringe is (3/2)λ.

Answer

Let's start by understanding the concept of path difference in Young's double slit experiment.

When a wavefront from one source (say, S1) travels to a point P on the screen, and another wavefront from the other source (S2) reaches the same point P, there will be a phase difference between the two wavefronts. This phase difference is called the path difference and is denoted by δ.

The path difference depends on the distance between the two sources (d), the distance between the sources and the screen (D), and the angle θ between the line joining the sources and the point P on the screen.

The path difference can be calculated using the formula:

δ = d sinθ

Now, let's use this formula to solve the given problem.

We are given that the wavelength of light used is 6000 Å (or 6 x 10^-7 m).

We are also given that the third dark fringe is formed at point P on the screen. This means that the path difference between the two wavefronts reaching point P is equal to (2n - 1) λ/2, where n is the order of the fringe.

For the third dark fringe, n = 3. Therefore, the path difference is:

δ = (2n - 1) λ/2 = (2 x 3 - 1) (6 x 10^-7)/2 = 5 x 10^-7 m

Now, we need to find the distance between source S1 and point P, which is the path travelled by the wavefront from source S1. Let's call this distance x.

Using trigonometry, we can write:

sinθ = x/D

Therefore,

x = D sinθ

We don't know the value of D or θ, but we can use the fact that the third dark fringe is formed at point P. This means that the path difference between the two wavefronts reaching P must be equal to the path difference between the two wavefronts at the second bright fringe, which is given by:

δ = n λ = 3 (6 x 10^-7) = 1.8 x 10^-6 m

Equating this to the path difference we calculated earlier, we get:

d sinθ = 1.8 x 10^-6 - 5 x 10^-7

d sinθ = 1.3 x 10^-6

Now, we need to find the value of d sinθ. We can use the fact that the third dark fringe is formed, which means that the angle θ is given by:

sinθ = (3/2) λ/d

Plugging in the values, we get:

sinθ = (3/2) (6 x 10^-7)/d

sinθ = 9 x 10^-7/d

Substituting this in the earlier equation, we get:

d (9 x 10^-7/d) = 1.3 x 10^-6

d cancels out, and we get:

sinθ = 1.44

This is not possible, as the maximum value of sinθ is 1. Therefore, our assumption that the third dark fringe is formed at point P must be wrong.

We can try again with a different fringe,

Similar JEE Doubts

In Youngs double slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000 coming from the coherent sources S1 and S2. At certain point P on the screen third dark fringe is formed. Then the path difference S1p − S2p in microns is

The figure shows the interference pattern obtained in a double-slit experiment using light of wavelength 600nm. 1, 2, 3, 4 and 5 are marked on five fringes.The third order bright fringe is

Top Courses for JEEView all

Physics for JEE Main & Advanced

Mock Tests for JEE Main and Advanced 2025

Mathematics (Maths) for JEE Main & Advanced

Chemistry for JEE Main & Advanced

Study Plan for JEE

Top Courses for JEE

Top Courses for JEE

Related Content

JEE Courses

Mock Test Series

Past Year Papers

Additional Courses

Company