To solve the problem step by step, we will follow the concepts of Young's Double Slit Experiment (YDSE) and the interference of light in a medium with a refractive index.
### Step 1: Understanding the Setup
In the YDSE setup, we have two slits (S1 and S2) and a screen where the interference pattern is observed. The experiment is conducted in a liquid with a refractive index \( \mu = 1.3 \), and there is a thin film of air in front of the lower slit (S2).
### Step 2: Wavelength in the Medium
The wavelength of light in the medium (liquid) can be calculated using the formula:
\[
\lambda = \frac{\lambda_0}{\mu}
\]
where \( \lambda_0 = 0.78 \, \mu m \) (wavelength in air) and \( \mu = 1.3 \).
Calculating this gives:
\[
\lambda = \frac{0.78 \, \mu m}{1.3} = 0.6 \, \mu m
\]
### Step 3: Condition for Maxima
For constructive interference (maxima), the condition is given by:
\[
t = \frac{(m + \frac{1}{2}) \lambda}{\mu}
\]
where \( t \) is the thickness of the air film, \( m \) is the order of the maxima, and \( \lambda \) is the wavelength in the medium.
### Step 4: Finding Thickness for Third Order Maxima
Given that the third order maximum is formed at the origin (O), we take \( m = 3 \):
\[
t = \frac{(3 + \frac{1}{2}) \cdot \lambda}{\mu} = \frac{(3.5) \cdot 0.6 \, \mu m}{1.3}
\]
Calculating this gives:
\[
t = \frac{2.1 \, \mu m}{1.3} \approx 1.615 \, \mu m
\]
### Step 5: Finding Position of Fourth Maxima
For the fourth order maximum (\( m = 4 \)):
\[
t = \frac{(4 + \frac{1}{2}) \cdot \lambda}{\mu} = \frac{(4.5) \cdot 0.6 \, \mu m}{1.3}
\]
Calculating this gives:
\[
t = \frac{2.7 \, \mu m}{1.3} \approx 2.077 \, \mu m
\]
### Step 6: Finding Position on the Screen
The position of the maxima on the screen can be calculated using the formula:
\[
y = \frac{m \cdot D \cdot \lambda}{d}
\]
where \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. Given \( \frac{D}{d} = 1000 \), we can express \( y \) for the fourth maximum:
\[
y_4 = \frac{4 \cdot D \cdot \lambda}{d}
\]
Substituting \( \lambda = 0.6 \, \mu m \):
\[
y_4 = \frac{4 \cdot 1000 \cdot 0.6 \, \mu m}{1} = 2400 \, \mu m = 2.4 \, mm
\]
### Final Results
1. The thickness of the air film for the third order maximum is approximately \( 1.615 \, \mu m \).
2. The position of the fourth maximum is approximately \( 2.4 \, mm \).
