What is the formula for calculating fringe width in an interference pattern?

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

When the system is in air

$d \sin {\theta} = n \lambda$ ...(1)

When the system is in water and slit separation is 2d,

$\mu 2d \sin {\theta} = n \lambda$ ...(2)

$\tan {\theta} = \frac y D$ ...( 3)

Taking d<<D, $\tan {\theta} \approx \sin {\theta}$ ...(4)

$\frac { n \lambda }{ 2 \mu d} = \frac { y} D$ ...(5)

$y = \frac { n \lambda D }{ 2 \mu d}$ ...( 6)

Fringe width $= \frac { \lambda D }{ 2 \mu d}$ ...(7)

Now, $\frac { \lambda D}{ d} =s$ ...( 8)

Fringe width $= \frac { s }{ 2 \mu } = \frac { 3s} 8$So, the correct option is (e).

Is this correct?

 

[ehild]

Correct. Nice work!

 

[Pushoam]

Correct. Nice work!

Is there any easier way to solve it?
Or one has to go through all of those steps for solving the above question.

 

[ehild]

Is there any easier way to solve it?
Or one has to go through all of those steps for solving the above question.

Was not it easy? You presented a nice solution, every step explained and clear.

 

[Pushoam]

Was not it easy? You presented a nice solution, every step explained and clear.

It was a multiple choice question. So, I was looking for a shorter way.

 

[ehild]

It was a multiple choice question. So, I was looking for a shorter way.

Most steps were explained during the class to you, so it was not needed to write down. You have the formula for the distance between nearest interference fringes, and it is proportional to lambda and inversely proportional to d. s' / s = λ'/λ d/d' The wavelength in a medium is the vacuum wavelength divided by the refractive index. So s'/s=(3/4) x (1/2 ).