Question

Young's double slit experiment is first performed in air and then in a medium other than air. It is found that the 8th bright fringe in the medium lies where the 5th dark fridge lies in air. The refractive index of the medium is nearly:
A. 1.78
B. 1.25
C. 1.59
D. 1.69

Answer 1

Hint: Obtain the formula used to find the position of the bright and dark fringes in a Young’s double slit experiment both in air and in the other medium. Find the position of the 5th dark fringe in air and the position of the 8th bright fringe in the medium. Equate them to find the refractive index of the medium.

Complete step-by-step solution
Young’s double slit uses two coherent sources of light, the interference pattern of which created these dark and bright fringes.
Position of nth dark fringe in Young’s double slit experiment is given by,
$x=(2n-1){\displaystyle \frac{\lambda D}{2d}}$
Position of nth bright fringe in Young's double slit experiment is given by,
$x={\displaystyle \frac{n\lambda D}{d}}$
If we perform the experiment in a medium the position of the nth bright fringe will be,
$x={\displaystyle \frac{n\lambda D}{\mu d}}$
Where, $\mu$ is the refractive index of the medium.
Now, position of 5th dark fringe in air is given by,
$\begin{array}{rl}& x_1=(2\times 5-1){\displaystyle \frac{\lambda D}{2d}}\\ & x_1={\displaystyle \frac{9\lambda D}{2d}}\end{array}$
Position of 8th bright fringe in medium is given by,
$x_2={\displaystyle \frac{8\lambda D}{\mu d}}$
Now in the question it is given that, the position of the 8th bright fringe in the medium lies where 5th dark fridge lies in air.
So,
$\begin{array}{rl}& x_1=x_2\\ & {\displaystyle \frac{9\lambda D}{2d}}={\displaystyle \frac{8\lambda D}{\mu d}}\\ & {\displaystyle \frac{9}{2}}={\displaystyle \frac{8}{\mu }}\\ & \mu =8\times {\displaystyle \frac{2}{9}}\\ & \mu ={\displaystyle \frac{16}{9}}\\ & \mu =1.777\\ & \mu \approx 1.78\end{array}$
So, the refractive index of the medium will be approximately 1.78.
The correct option is (A).

Note: We can also find the solution of this question by taking the wavelength in both mediums as different. Then we can just put the values in the direct formulas and then take the ratio of the wavelength of light in the two media which will be our refractive index for the given question.