A thin of liquid polymer, `n = 1.25`, coats a slab of Pyrex, `n = 1.50`. White light is incident perpendicularly on the film. In the reflection, full destructive interference occurs for `lambda = 600 nm` and full constructive interference occurs for `lambda = 700 nm` What is the thickness of the polymer film?

Options

A. 120 nm
B. 280 nm
C. 460 nm
D. 840 nm

Explanation

To find the thickness of the polymer film that causes full destructive interference for a wavelength of 600 nm and full constructive interference for a wavelength of 700 nm, we can follow these steps:

### Step 1: Understanding the Conditions for Interference
For thin films, the conditions for interference depend on the phase changes that occur upon reflection. When light reflects off a medium of higher refractive index, it undergoes a phase change of π (or half a wavelength).

- **Destructive Interference** (for λ = 600 nm):
The condition for destructive interference is given by:
\[ 2nt = (m + \frac{1}{2})\lambda \]
where \( n \) is the refractive index of the film, \( t \) is the thickness of the film, \( m \) is an integer (order of interference), and \( \lambda \) is the wavelength of light in vacuum.

- **Constructive Interference** (for λ = 700 nm):
The condition for constructive interference is given by:
\[ 2nt = m\lambda \]

### Step 2: Setting Up the Equations
Let’s denote:
- \( n = 1.25 \) (refractive index of the polymer)
- \( t \) = thickness of the polymer film
- \( \lambda_1 = 600 \, \text{nm} \)
- \( \lambda_2 = 700 \, \text{nm} \)

From the destructive interference condition:
\[ 2(1.25)t = (m_1 + \frac{1}{2})600 \]
From the constructive interference condition:
\[ 2(1.25)t = m_2 \cdot 700 \]

### Step 3: Equating the Two Conditions
Since both equations equal \( 2nt \), we can set them equal to each other:
\[ (m_1 + \frac{1}{2})600 = m_2 \cdot 700 \]

### Step 4: Solving for Integer Values of m
We can express \( m_2 \) in terms of \( m_1 \):
\[ m_2 = \frac{(m_1 + \frac{1}{2})600}{700} \]
This simplifies to:
\[ m_2 = \frac{600m_1 + 300}{700} \]
To find integer solutions, we can try different integer values for \( m_1 \).

### Step 5: Testing Integer Values
Let’s try \( m_1 = 3 \):
\[ m_2 = \frac{600 \cdot 3 + 300}{700} = \frac{1800 + 300}{700} = \frac{2100}{700} = 3 \]
Thus, \( m_1 = 3 \) and \( m_2 = 3 \) works.

### Step 6: Finding Thickness
Now, we can substitute \( m_1 \) back into either equation to find \( t \):
Using the destructive interference equation:
\[ 2(1.25)t = (3 + \frac{1}{2})600 \]
\[ 2.5t = 3.5 \cdot 600 \]
\[ 2.5t = 2100 \]
\[ t = \frac{2100}{2.5} = 840 \, \text{nm} \]

### Final Answer
The thickness of the polymer film is:
\[ \boxed{840 \, \text{nm}} \]