Question

37)
Calculate the separation between 3rd bright and 5th dark fringe of an interference pattern obtained in
a young's double slit experiment. The double slit separation is 1mm and the screen is 1.5m away.
Wavelength of light used is 550nm. What is the fringe width when the screen is moved 0.5 m away
from the plane of the double slit?​

Answer 1

Answer:

sorry I didn't know that

Answer 2

Given:

Given that,

$\[d=1mm=1\times {{10}^{-3}}m\]$

$\[D=1.5m\]$

$\[\lambda =550nm=550\times {{10}^{-9}}m\]$

$\[r=1.5m\]$

To find:

We need to find what is the fringe width when the screen is moved 0.5 m away from the plane of the double slit.

Solution:

Generally, Fringe width formula is $\[X=\frac{\lambda D}{d}\]$.

Apply those values within the above equation, we have to get fringe width of slits.

$& X=\frac{550\times {{10}^{-9}}\times 1.5}{1\times {{10}^{-3}}} \\ & \\ & X=550\times 1.5\times {{10}^{-6}} \\ & \\ & X=825\times {{10}^{-6}} \\ & \\ & X=0.82\times {{10}^{-3}}m$

When the screen is moved 0.5 m away from the plane of the double slit, we have to calculate the value of fringe width.

According to the problem the distance between the slits is given and let us considers it is d and the width of the fringe is also given and we will consider it as s.

        Let λ be the wavelength.

        Let r be the distance between the slit and the screen.

The fringe width  $\[s=\frac{r\lambda }{d}\]$

$& s=\frac{\left( 1.5+0.5 \right)550\times {{10}^{-9}}}{1\times {{10}^{-3}}} \\ & \\ & s=2\times 550\times {{10}^{-9}}\times {{10}^{3}} \\ & \\ & s=1100\times {{10}^{-6}} \\ & \\ & s=11\times {{10}^{2}}\times {{10}^{-6}} \\ & \\ & s=11\times {{10}^{-4}} \\ & \\ & s=0.11\times {{10}^{-2}}m \\ & \\ & s=0.11cm$

So, the fringe width is $\[s=0.11cm\]$. When the screen is moved 0.5 m away from the plane of the double slit.