Question

Imagine & YDSE set-up immersed in a medium with refractive index p. What will the ratio of the new path difference and the fridge width, respectively, with the old path difference

Answer 1

1, 1/μ

Answer 2

When a YDSE setup is immersed in a medium with refractive index $\mu$, the wavelength of the light changes. Let the wavelength of light in vacuum (or air) be $\lambda_{old}$. The speed of light in vacuum is $c$. The frequency of light is $f = c/\lambda_{old}$.

When light enters a medium with refractive index $\mu$, the speed of light in the medium is $v = c/\mu$. The frequency of light remains the same, $f$. The wavelength of light in the medium, $\lambda_{new}$, is given by $v = f \lambda_{new}$. Substituting the values, we get $c/\mu = (c/\lambda_{old}) \lambda_{new}$. So, the new wavelength in the medium is $\lambda_{new} = \frac{\lambda_{old}}{\mu}$.

The path difference between the waves from the two slits reaching a point on the screen is a geometrical path difference. For a point at a distance $y$ from the center of the screen, the path difference is given by $\Delta x = \frac{yd}{D}$ (for small angles), where $d$ is the distance between the slits and $D$ is the distance between the slits and the screen. This geometrical path difference depends only on the geometry of the setup ($y$, $d$, $D$). When the YDSE setup is immersed in a medium, the geometry remains unchanged. Therefore, the path difference (geometrical) at any point on the screen remains the same. Let $\Delta x_{old}$ be the path difference in the original medium (air/vacuum) and $\Delta x_{new}$ be the path difference in the new medium. $\Delta x_{new} = \Delta x_{old}$. The ratio of the new path difference to the old path difference is $\frac{\Delta x_{new}}{\Delta x_{old}} = \frac{\Delta x_{old}}{\Delta x_{old}} = 1$.

The fringe width in a YDSE setup is given by the formula $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength of light, $D$ is the distance between the slits and the screen, and $d$ is the distance between the slits. In the original medium (air/vacuum), the fringe width is $\beta_{old} = \frac{\lambda_{old} D}{d}$. In the new medium with refractive index $\mu$, the wavelength of light is $\lambda_{new} = \frac{\lambda_{old}}{\mu}$. The distances $D$ and $d$ remain unchanged. The new fringe width is $\beta_{new} = \frac{\lambda_{new} D}{d}$. Substitute the value of $\lambda_{new}$: $\beta_{new} = \frac{(\lambda_{old}/\mu) D}{d} = \frac{1}{\mu} \frac{\lambda_{old} D}{d}$. So, $\beta_{new} = \frac{\beta_{old}}{\mu}$. The ratio of the new fringe width to the old fringe width is $\frac{\beta_{new}}{\beta_{old}} = \frac{\beta_{old}/\mu}{\beta_{old}} = \frac{1}{\mu}$.

The question asks for the ratio of the new path difference and the fringe width, respectively, with the old path difference. Based on the common phrasing in physics problems, this is interpreted as asking for:

1. The ratio of the new path difference to the old path difference.
2. The ratio of the new fringe width to the old fringe width.

Ratio 1: $\frac{\text{New path difference}}{\text{Old path difference}} = \frac{\Delta x_{new}}{\Delta x_{old}} = 1$. Ratio 2: $\frac{\text{New fringe width}}{\text{Old fringe width}} = \frac{\beta_{new}}{\beta_{old}} = \frac{1}{\mu}$.

So the required ratios are 1 and $1/\mu$, respectively.

Explanation of the solution:

1. The geometrical path difference at any point in YDSE depends on the geometry of the setup ($d$, $D$, $y$) and not on the medium. Thus, the new path difference is equal to the old path difference, making their ratio 1.
2. When the YDSE setup is immersed in a medium with refractive index $\mu$, the wavelength of light changes from $\lambda_{old}$ to $\lambda_{new} = \lambda_{old}/\mu$.
3. The fringe width is proportional to the wavelength ($\beta = \lambda D/d$). Since the wavelength becomes $\lambda_{old}/\mu$, the new fringe width becomes $\beta_{new} = (\lambda_{old}/\mu) D/d = \beta_{old}/\mu$.
4. The ratio of the new fringe width to the old fringe width is $\beta_{new}/\beta_{old} = ( \beta_{old}/\mu ) / \beta_{old} = 1/\mu$.
5. The question asks for the ratio of the new path difference and the fringe width, respectively, with the old path difference. This is interpreted as the ratio of new path difference to old path difference, and the ratio of new fringe width to old fringe width. These ratios are 1 and $1/\mu$, respectively.