White light may be considered to be mixture of waves of `lambda` ranging between `3900 Å` and `7800 Å`. An oil film of thickness `10,000 Å` is examined normally by the reflected light. If `mu=1.4`, then the film appears bright for

Options

A. 4308Å, 5091Å,6222Å
B. 4000Å,5091Å,5600Å
C. 4667Å,6222Å,7000Å
D. 4000Å,4667Å,5600Å,7000Å

Answer

To determine the wavelengths for which the oil film appears bright, we will use the condition for constructive interference in thin films. Here’s a step-by-step solution:

### Step 1: Understanding the Condition for Constructive Interference
For constructive interference in a thin film, the condition is given by:
\[ 2\mu t = (n + \frac{1}{2}) \lambda \]
where:
- \( \mu \) is the refractive index of the film,
- \( t \) is the thickness of the film,
- \( n \) is the order of interference (0, 1, 2, ...),
- \( \lambda \) is the wavelength of light in vacuum.

### Step 2: Rearranging the Formula
We can rearrange the formula to solve for \( \lambda \):
\[ \lambda = \frac{2\mu t}{n + \frac{1}{2}} \]

### Step 3: Substituting Known Values
Given:
- \( \mu = 1.4 \)
- \( t = 10,000 \, \text{Å} = 10,000 \times 10^{-10} \, \text{m} = 10^{-6} \, \text{m} \)

Now substituting these values into the equation:
\[ \lambda = \frac{2 \times 1.4 \times 10,000}{n + \frac{1}{2}} \]
\[ \lambda = \frac{28,000}{n + \frac{1}{2}} \]

### Step 4: Finding Wavelengths for Different Orders of Interference
Now we will calculate \( \lambda \) for different values of \( n \):

1. For \( n = 0 \):
\[ \lambda = \frac{28,000}{0 + \frac{1}{2}} = \frac{28,000}{0.5} = 56,000 \, \text{Å} \]

2. For \( n = 1 \):
\[ \lambda = \frac{28,000}{1 + \frac{1}{2}} = \frac{28,000}{1.5} \approx 18,667 \, \text{Å} \]

3. For \( n = 2 \):
\[ \lambda = \frac{28,000}{2 + \frac{1}{2}} = \frac{28,000}{2.5} = 11,200 \, \text{Å} \]

4. For \( n = 3 \):
\[ \lambda = \frac{28,000}{3 + \frac{1}{2}} = \frac{28,000}{3.5} \approx 8,000 \, \text{Å} \]

5. For \( n = 4 \):
\[ \lambda = \frac{28,000}{4 + \frac{1}{2}} = \frac{28,000}{4.5} \approx 6,222 \, \text{Å} \]

6. For \( n = 5 \):
\[ \lambda = \frac{28,000}{5 + \frac{1}{2}} = \frac{28,000}{5.5} \approx 5,091 \, \text{Å} \]

7. For \( n = 6 \):
\[ \lambda = \frac{28,000}{6 + \frac{1}{2}} = \frac{28,000}{6.5} \approx 4,307 \, \text{Å} \]

### Step 5: Identifying Valid Wavelengths
The wavelengths calculated must fall within the visible spectrum of white light, which ranges from \( 3900 \, \text{Å} \) to \( 7800 \, \text{Å} \). The valid wavelengths from our calculations are:
- For \( n = 0 \): \( 56,000 \, \text{Å} \) (not valid)
- For \( n = 1 \): \( 18,667 \, \text{Å} \) (not valid)
- For \( n = 2 \): \( 11,200 \, \text{Å} \) (not valid)
- For \( n = 3 \): \( 8,000 \, \text{Å} \) (not valid)
- For \( n = 4 \): \( 6,222 \, \text{Å} \) (valid)
- For \( n = 5 \): \( 5,091 \, \text{Å} \) (valid)
- For \( n = 6 \): \( 4,307 \, \text{Å} \) (valid)

### Conclusion
The oil film appears bright for wavelengths \( 6,222 \, \text{Å} \), \( 5,091 \, \text{Å} \), and \( 4,307 \, \text{Å} \).