White light reflected at normal incidence from a soap film has minima at `6500Å` and maxima at `7500Å` in the visible region without minimum in between. If `mu` is `(5)/(3)` for the thin film, thickness of the film is

Options

A. `7.40xx10^(-7)m`
B. `9.75xx10^(-5)mm`
C. `9.40xx10^(-7)cm`
D. `9.75xx10^(-4)mm`

Answer

To find the thickness of the soap film, we will use the conditions for constructive and destructive interference in thin films. Here’s the step-by-step solution:

### Step 1: Understand the Conditions for Minima and Maxima
For a thin film, the conditions for maxima and minima are given by:
- For maxima: \( 2 \mu t = (n + \frac{1}{2}) \lambda_{\text{max}} \)
- For minima: \( 2 \mu t = n \lambda_{\text{min}} \)

Where:
- \( \mu \) is the refractive index of the film,
- \( t \) is the thickness of the film,
- \( n \) is an integer (order of interference),
- \( \lambda_{\text{max}} \) and \( \lambda_{\text{min}} \) are the wavelengths of light corresponding to maxima and minima respectively.

### Step 2: Set Up the Equations
From the problem:
- \( \lambda_{\text{min}} = 6500 \, \text{Å} = 6500 \times 10^{-10} \, \text{m} \)
- \( \lambda_{\text{max}} = 7500 \, \text{Å} = 7500 \times 10^{-10} \, \text{m} \)
- \( \mu = \frac{5}{3} \)

We can write the equations for maxima and minima:
1. \( 2 \mu t = (n + \frac{1}{2}) \lambda_{\text{max}} \)
2. \( 2 \mu t = n \lambda_{\text{min}} \)

### Step 3: Equate the Two Expressions
Since both expressions are equal to \( 2 \mu t \), we can set them equal to each other:
\[ (n + \frac{1}{2}) \lambda_{\text{max}} = n \lambda_{\text{min}} \]

### Step 4: Substitute the Values
Substituting the values of \( \lambda_{\text{max}} \) and \( \lambda_{\text{min}} \):
\[ (n + \frac{1}{2}) (7500 \times 10^{-10}) = n (6500 \times 10^{-10}) \]

### Step 5: Simplify the Equation
Cancelling \( 10^{-10} \) from both sides:
\[ (n + \frac{1}{2}) (7500) = n (6500) \]
Expanding this gives:
\[ 7500n + 3750 = 6500n \]

### Step 6: Solve for \( n \)
Rearranging the equation:
\[ 7500n - 6500n = -3750 \]
\[ 1000n = -3750 \]
\[ n = \frac{3750}{1000} = 3.75 \]

### Step 7: Use \( n \) to Find Thickness \( t \)
Now we can use either equation to find \( t \). Let’s use the minima equation:
\[ 2 \mu t = n \lambda_{\text{min}} \]
Substituting \( \mu \) and \( n \):
\[ 2 \left(\frac{5}{3}\right) t = 3.75 \times (6500 \times 10^{-10}) \]
\[ \frac{10}{3} t = 3.75 \times 6500 \times 10^{-10} \]

### Step 8: Calculate \( t \)
Calculating the right side:
\[ t = \frac{3.75 \times 6500 \times 10^{-10} \times 3}{10} \]
Calculating the numerical value:
\[ t = \frac{24375 \times 10^{-10} \times 3}{10} = 7.3125 \times 10^{-7} \, \text{m} = 7.3125 \times 10^{-7} \, \text{m} \]

### Final Answer
The thickness of the film is approximately:
\[ t \approx 7.31 \times 10^{-7} \, \text{m} \]