White light reflected from a soap film (Refractive index `=1.5` ) has a maxima at 600 nm and a minima at 450 nm at with no minimum in between. Then, the thickness of the film is

Options

A. 1 micro meter
B. 2 micro meter
C. 3 micro meter
D. 4 micro meter

Explanation

To solve the problem of finding the thickness of a soap film given the maxima and minima of reflected white light, we can follow these steps:

### Step 1: Understand the Conditions for Maxima and Minima
For a soap film with a refractive index \( \mu = 1.5 \):
- The condition for maxima (constructive interference) is given by:
\[ 2 \mu t \cos r = \left(2n + 1\right) \frac{\lambda_1}{2} \]
- The condition for minima (destructive interference) is given by:
\[ 2 \mu t \cos r = n \lambda_2 \]

Where:
- \( t \) is the thickness of the film,
- \( r \) is the angle of refraction (which we can assume to be 0 for normal incidence, making \( \cos r = 1 \)),
- \( \lambda_1 = 600 \, \text{nm} \) (for maxima),
- \( \lambda_2 = 450 \, \text{nm} \) (for minima),
- \( n \) is an integer representing the order of interference.

### Step 2: Set Up the Equations
Using the values given:
1. For maxima at \( \lambda_1 = 600 \, \text{nm} \):
\[ 2 \times 1.5 \times t = (2n + 1) \frac{600 \, \text{nm}}{2} \]
Simplifying this gives:
\[ 3t = (2n + 1) \times 300 \]
(Equation 1)

2. For minima at \( \lambda_2 = 450 \, \text{nm} \):
\[ 2 \times 1.5 \times t = n \times 450 \, \text{nm} \]
Simplifying this gives:
\[ 3t = 450n \]
(Equation 2)

### Step 3: Divide the Equations
Now, we can divide Equation 1 by Equation 2 to eliminate \( t \):
\[ \frac{(2n + 1) \times 300}{450n} = 1 \]
Cross-multiplying gives:
\[ (2n + 1) \times 300 = 450n \]
Expanding this:
\[ 600n + 300 = 450n \]
Rearranging gives:
\[ 600n - 450n = -300 \]
\[ 150n = -300 \]
\[ n = 2 \]

### Step 4: Substitute Back to Find Thickness
Now that we have \( n = 2 \), we can substitute this back into either Equation 1 or Equation 2 to find \( t \). Using Equation 1:
\[ 3t = (2 \times 2 + 1) \times 300 \]
\[ 3t = (4 + 1) \times 300 \]
\[ 3t = 5 \times 300 \]
\[ 3t = 1500 \]
\[ t = \frac{1500}{3} = 500 \, \text{nm} \]

### Final Result
Thus, the thickness of the soap film is:
\[ t = 500 \, \text{nm} \quad \text{or} \quad 5 \times 10^{-7} \, \text{m} \]