Question

Young's double slit experiment is first done in air and then in a medium of refractive index $\mu$. If the 7th dark fringe in the medium lies where the 4th bright fringe is in air, then the value of $\mu$ is:

Answer 1

1.625

Answer 2

The position of the $n^{th}$ bright fringe in a Young's Double Slit Experiment (YDSE) is given by:

$y_{n, bright} = \frac{n \lambda D}{d}$

where $n$ is the order of the bright fringe (1, 2, 3, ...), $\lambda$ is the wavelength of light, $D$ is the distance between the slits and the screen, and $d$ is the distance between the two slits.

The position of the $n^{th}$ dark fringe is given by:

$y_{n, dark} = \frac{(2n-1) \lambda D}{2d}$

where $n$ is the order of the dark fringe (1, 2, 3, ...).

When the experiment is performed in a medium with refractive index $\mu$, the wavelength of light changes. If $\lambda_a$ is the wavelength of light in air, then the wavelength of light in the medium, $\lambda_m$, is given by:

$\lambda_m = \frac{\lambda_a}{\mu}$

According to the problem statement:

The 7th dark fringe in the medium lies where the 4th bright fringe is in air. So, $y_{7, dark, medium} = y_{4, bright, air}$.

Let's write down the expressions for these positions:

Position of the 4th bright fringe in air:

$y_{4, bright, air} = \frac{4 \lambda_a D}{d}$

Position of the 7th dark fringe in the medium:

For the 7th dark fringe, $n=7$.

$y_{7, dark, medium} = \frac{(2 \times 7 - 1) \lambda_m D}{2d} = \frac{(14 - 1) \lambda_m D}{2d} = \frac{13 \lambda_m D}{2d}$

Now, equate the two positions:

$\frac{13 \lambda_m D}{2d} = \frac{4 \lambda_a D}{d}$

We can cancel out the common terms $D$ and $d$ from both sides:

$\frac{13 \lambda_m}{2} = 4 \lambda_a$

Now, substitute $\lambda_m = \frac{\lambda_a}{\mu}$:

$\frac{13}{2} \left(\frac{\lambda_a}{\mu}\right) = 4 \lambda_a$

Cancel out $\lambda_a$ from both sides:

$\frac{13}{2\mu} = 4$

Now, solve for $\mu$:

$13 = 4 \times 2\mu$ $13 = 8\mu$ $\mu = \frac{13}{8}$

$\mu = 1.625$