What is the minimum thickness of a thin film `(mu=1.2)` that results in constructive interference in the reflected light? If the film is illuminated with light whose wavelength in free space is `lambda=500nm` ?

Options

A. 104 nm
B. 200 nm
C. 300 nm
D. 400 nm

Explanation

To find the minimum thickness of a thin film that results in constructive interference in the reflected light, we can follow these steps:

### Step 1: Understand the Condition for Constructive Interference
For a thin film with a refractive index \( \mu \), the condition for constructive interference in reflected light (when the film is illuminated from air) is given by the formula:
\[ 2 \mu t = \left(n - \frac{1}{2}\right) \lambda \]
where \( n \) is an integer (1, 2, 3, ...), \( t \) is the thickness of the film, and \( \lambda \) is the wavelength of light in vacuum.

### Step 2: Set Up for Minimum Thickness
To find the minimum thickness, we set \( n = 1 \):
\[ 2 \mu t = \left(1 - \frac{1}{2}\right) \lambda \]
This simplifies to:
\[ 2 \mu t = \frac{1}{2} \lambda \]

### Step 3: Solve for Thickness \( t \)
Now, we can rearrange the equation to solve for \( t \):
\[ t = \frac{\lambda}{4 \mu} \]

### Step 4: Substitute Given Values
We know:
- \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- \( \mu = 1.2 \)

Substituting these values into the equation:
\[ t = \frac{500 \times 10^{-9}}{4 \times 1.2} \]

### Step 5: Calculate the Thickness
Calculating the denominator:
\[ 4 \times 1.2 = 4.8 \]
Now substituting back:
\[ t = \frac{500 \times 10^{-9}}{4.8} \approx 104.17 \times 10^{-9} \, \text{m} \]
Converting to nanometers:
\[ t \approx 104.17 \, \text{nm} \]

### Step 6: Round the Result
Rounding to the nearest whole number, we get:
\[ t \approx 104 \, \text{nm} \]

### Final Answer
The minimum thickness of the thin film that results in constructive interference in the reflected light is approximately **104 nm**.

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