Double-Slit Experiment Setup and Initial Data

This problem involves analyzing a double-slit experiment. We are given information about the light's wavelength ($\lambda$), the positions of fringes like the central maximum and the second bright fringe, the distance to the screen (D), and the refractive index (n) of a liquid the setup is immersed in. The goal is to find the angular separation between the first and fourth dark fringes in the liquid.

In a double-slit experiment, the positions of the fringes are determined by the wavelength of light ($\lambda$), the distance between the slits (d), and the distance to the screen (D).

• The linear position ($y_n$) of the n-th bright fringe is described by the formula: $y_n = \frac{n \lambda D}{d}$
• The linear position ($y'_m$) of the m-th dark fringe is described by: $y'_m = \frac{(m + 1/2) \lambda D}{d}$
• For small angles, the angular position ($\theta$) is approximately $\theta \approx \frac{y}{D}$. Therefore, the angular position of the n-th bright fringe is $\theta_n = \frac{n \lambda}{d}$, and the angular position of the m-th dark fringe is $\theta'_m = \frac{(m + 1/2) \lambda}{d}$.

We will use these relationships to calculate the slit separation (d) from the initial conditions in air and then determine the required angular separation in the liquid.

Slit Separation Determination

First, we need to determine the distance between the two slits (d) using the experiment's initial setup in air.

• Initial wavelength in air ($\lambda_{air}$) = $\text{600 nm} = 600 \times 10^{-9}$ m.
• Distance to the screen (D) = $\text{1.5 m}$.
• The linear distance ($y_2$) between the central maximum (n=0) and the second bright fringe (n=2) is $\text{3 mm} = 3 \times 10^{-3}$ m.

Using the formula for the position of a bright fringe ($y_n = \frac{n \lambda_{air} D}{d}$), we can write for the second bright fringe (n=2):

$y_2 = \frac{2 \lambda_{air} D}{d}$

Rearranging the formula to solve for the slit separation (d):

$d = \frac{2 \lambda_{air} D}{y_2}$

Substituting the given values:

$d = \frac{2 \times (600 \times 10^{-9} \text{ m}) \times (1.5 \text{ m})}{3 \times 10^{-3} \text{ m}}$

$d = \frac{1800 \times 10^{-9}}{3 \times 10^{-3}} \text{ m}$

$d = 600 \times 10^{-6} \text{ m}$

Thus, the slit separation d is $600 \times 10^{-6}$ meters (or $0.6$ mm).

Wavelength in Liquid Calculation

Next, we calculate the wavelength of the light when it travels through the liquid. The wavelength changes based on the medium's refractive index (n).

• Refractive index of the liquid (n) = $\text{1.5}$.
• Wavelength in air ($\lambda_{air}$) = $\text{600 nm}$.

The relationship between the wavelength in air and the wavelength in the liquid ($\lambda_{liquid}$) is:

$\lambda_{liquid} = \frac{\lambda_{air}}{n}$

Substituting the values:

$\lambda_{liquid} = \frac{600 \text{ nm}}{1.5}$

$\lambda_{liquid} = \text{400 nm} = 400 \times 10^{-9}$ m

So, the effective wavelength of light inside the liquid is $\text{400 nm}$.

Angular Separation of First and Fourth Dark Fringes

Now, we calculate the angular separation between the first and fourth dark fringes in the liquid. The angular position of the m-th dark fringe is given by $\theta'_m = \frac{(m + 1/2) \lambda_{liquid}}{d}$.

• The first dark fringe occurs when $m=0$.
• The fourth dark fringe occurs when $m=3$.

The angular position of the first dark fringe ($m=0$) is:

$\theta'_{0} = \frac{(0 + 1/2) \lambda_{liquid}}{d} = \frac{0.5 \lambda_{liquid}}{d}$

The angular position of the fourth dark fringe ($m=3$) is:

$\theta'_{3} = \frac{(3 + 1/2) \lambda_{liquid}}{d} = \frac{3.5 \lambda_{liquid}}{d}$

The required angular separation ($\Delta \theta$) is the difference between these two angular positions:

$\Delta \theta = \theta'_{3} - \theta'_{0} = \frac{3.5 \lambda_{liquid}}{d} - \frac{0.5 \lambda_{liquid}}{d}$

$\Delta \theta = \frac{(3.5 - 0.5) \lambda_{liquid}}{d} = \frac{3.0 \lambda_{liquid}}{d}$

Substitute the calculated values of $\lambda_{liquid}$ and d:

$\Delta \theta = \frac{3.0 \times (400 \times 10^{-9} \text{ m})}{600 \times 10^{-6} \text{ m}}$

$\Delta \theta = \frac{1200 \times 10^{-9}}{600 \times 10^{-6}}$

$\Delta \theta = 2 \times 10^{-3}$ radians

Final Angular Separation in Degrees

The calculated angular separation is in radians. We need to convert this to degrees to match the options provided.

The conversion formula is:

Angle in degrees = Angle in radians $\times \frac{180}{\pi}$

Performing the conversion:

$\Delta \theta (\text{degrees}) = (2 \times 10^{-3}) \times \frac{180}{\pi}$

$\Delta \theta (\text{degrees}) = \frac{0.36}{\pi}$

Using the approximate value of $\pi \approx 3.14159$:

$\Delta \theta (\text{degrees}) \approx \frac{0.36}{3.14159} \approx 0.11459^\circ$

Rounding to four decimal places, the angular separation between the first and fourth dark fringes in the liquid is approximately $\text{0.1146}^\circ$.