Question

In Young’s double slits experiment a student observes the distance between two slits is 2mm and

the screen is placed at a distance 1 m from the slits. He also observes the distance between two

consecutive bright fringes is 0.3mm. He wants to increase the distance between two consecutive

fringes for this he reduces half of the distance between two slits and increases 50% of the distance of

the screen from slits.

Calculate the phase difference of 2nd bright fringe.


Does the student get success to the experiment? Analyze mathematically according

to the stem.​

Answer 1

Answer:

The joule ( jawl, jool; symbol: J) is a derived unit of energy in the International System of Units. It is equal to the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one metre (1 newton metre or N⋅m).

Explanation:

पागल

Answer 2

Answer:

Distance between two consecutive fringes can be increased by reducing the slit distance and increasing the screen distance.  

The phase difference of 2nd bright fringe is 4π.

Explanation:

Given the distance between two slits,  $d= 2mm=2\times 10^{-3} m$

The distance between the slit and the screen,  $D= 1 m$

The distance between two consecutive bright fringes is 0.3mm.

In Young's Double slit experiment, the separation between two consecutive bright fringes is given by

        $\beta=\dfrac{\lambda D}{d}$

where λ is the wavelength of incident light, D is the distance between the screen and slits and d is slit separation.

Substituting the given values,

$0.3\times 10^{-3} =\dfrac{\lambda \times 1}{2\times 10^{-3}}$

      $\implies \lambda =0.3\times 10^{-3}\times2\times 10^{-3} =0.3\times 10^{-6}m$

If the distance between two slits is reduced to half, i.e.,

$d= \dfrac{2mm}{2} = 10^{-3} m$

The distance between the slit and the screen is increased by 50%, then

$D=1+\dfrac{50}{100}=1.5m$

Then the separation between two bright fringes is

$\beta=\dfrac{\lambda D}{d} =\dfrac{0.3\times 10^{-3} \times 1.5}{10^{-3}}=0.45m$

So, the distance between two consecutive fringes is increased by reducing the slit distance and increasing the distance between the screen and the slit.  

So, the student get success to the experiment.

The path difference for the bright is given by $\Delta x=n\lambda$

where n is the order of the fringe.

And the phase difference is given by

$\Delta \phi=\dfrac{2\pi }{\lambda} \times \Delta x=\dfrac{2\pi }{\lambda} \times n\lambda=2\pi n$

Thus, the phase difference of 2nd bright fringe is

 $\Delta \phi=2\pi n=2\pi\times 2=4\pi$

Therefore, the phase difference of 2nd bright fringe is 4π.

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