A parallel beam of white light falls on a thin film whose refractive index is equal to `4//3`. The angle of incidence `i=53^@`. What must be the minimum film thickness if the reflected light is to be coloured yellow most intensively? `(tan53^@ =4//3)`

Explanation

To find the minimum film thickness required for the reflected light to be colored yellow most intensively, we can follow these steps:

### Step 1: Understand the Setup
We have a thin film with a refractive index \( \mu = \frac{4}{3} \) and a beam of white light incident at an angle \( i = 53^\circ \). We need to find the minimum thickness \( t \) of the film for constructive interference of yellow light.

### Step 2: Determine the Angle of Refraction
Using Snell's law:
\[ \mu_1 \sin i = \mu_2 \sin r \]
where \( \mu_1 = 1 \) (air), \( \mu_2 = \frac{4}{3} \), and \( i = 53^\circ \).

Substituting the values:
\[ 1 \cdot \sin(53^\circ) = \frac{4}{3} \sin(r) \]
\[ \sin(r) = \frac{3}{4} \sin(53^\circ) \]
Calculating \( \sin(53^\circ) \):
\[ \sin(53^\circ) \approx 0.7986 \quad \text{(using a calculator)} \]
Thus,
\[ \sin(r) = \frac{3}{4} \cdot 0.7986 \approx 0.59895 \]
Now, we find \( r \) using the inverse sine function:
\[ r \approx \sin^{-1}(0.59895) \approx 37^\circ \]

### Step 3: Calculate the Path Difference
The path difference \( \Delta x \) between the two rays (one reflected and one refracted) can be calculated as:
\[ \Delta x = 2t \mu \cos(r) - 2t \sin(i) \]

### Step 4: Express the Path Difference
Using the values:
- \( \mu = \frac{4}{3} \)
- \( \cos(37^\circ) \approx 0.7986 \)
- \( \sin(53^\circ) \approx 0.7986 \)

We can express the path difference:
\[ \Delta x = 2t \left(\frac{4}{3} \cdot 0.7986\right) - 2t \cdot 0.7986 \]
\[ \Delta x = 2t \left(\frac{4}{3} \cdot 0.7986 - 0.7986\right) \]
\[ \Delta x = 2t \cdot 0.7986 \left(\frac{4}{3} - 1\right) \]
\[ \Delta x = 2t \cdot 0.7986 \cdot \frac{1}{3} = \frac{2t \cdot 0.7986}{3} \]

### Step 5: Condition for Constructive Interference
For constructive interference, the path difference must equal \( n \lambda \) (where \( n \) is an integer and \( \lambda \) is the wavelength of yellow light). The minimum thickness corresponds to \( n = 1 \):
\[ \Delta x = \lambda \]

### Step 6: Substitute Wavelength for Yellow Light
Using \( \lambda = 0.6 \, \mu m = 0.6 \times 10^{-6} \, m \):
\[ \frac{2t \cdot 0.7986}{3} = 0.6 \times 10^{-6} \]
Solving for \( t \):
\[ t = \frac{3 \cdot 0.6 \times 10^{-6}}{2 \cdot 0.7986} \approx \frac{1.8 \times 10^{-6}}{1.5972} \approx 1.125 \times 10^{-6} \, m \]

### Step 7: Final Calculation
Converting to micrometers:
\[ t \approx 1.125 \, \mu m \]

### Conclusion
The minimum film thickness \( t \) required for the reflected light to be colored yellow most intensively is approximately \( 1.125 \, \mu m \).