A parallel beam of white light is incident on a thin film of air of uniform thickness. Wavlength `7200 Å` and `5400 Å` are observed to be missing from the spectrum of reflected light viewed normally. The other wavelength in the visible region missing in the reflected spectrum is

Options

A. `6000 Å`
B. `4320 Å`
C. `5500 Å`
D. `6500 Å`

Explanation

To solve the problem, we need to find the third missing wavelength in the reflected spectrum when white light is incident on a thin film of air. The wavelengths given are 7200 Å and 5400 Å. Let's go through the solution step by step.

### Step 1: Understanding the Condition for Destructive Interference
In a thin film, destructive interference occurs when the path difference between the light reflected from the top and bottom surfaces of the film leads to a phase difference of an odd multiple of π (or half a wavelength). The condition for destructive interference can be expressed as:
\[ 2 \mu t = n \lambda \]
where:
- \( \mu \) is the refractive index of the film (for air, \( \mu \approx 1 \)),
- \( t \) is the thickness of the film,
- \( n \) is an integer (order of interference),
- \( \lambda \) is the wavelength of light.

### Step 2: Setting Up the Equations
Given that two wavelengths are missing (7200 Å and 5400 Å), we can set up the equations for these wavelengths:
1. For \( \lambda_1 = 7200 \, \text{Å} \):
\[ 2 \mu t = n \cdot 7200 \]
2. For \( \lambda_2 = 5400 \, \text{Å} \):
\[ 2 \mu t = (n + 1) \cdot 5400 \]

### Step 3: Equating the Two Expressions
Since both expressions equal \( 2 \mu t \), we can set them equal to each other:
\[ n \cdot 7200 = (n + 1) \cdot 5400 \]

### Step 4: Solving for \( n \)
Expanding the equation:
\[ n \cdot 7200 = n \cdot 5400 + 5400 \]
Rearranging gives:
\[ n \cdot 7200 - n \cdot 5400 = 5400 \]
Factoring out \( n \):
\[ n(7200 - 5400) = 5400 \]
\[ n \cdot 1800 = 5400 \]
Thus,
\[ n = \frac{5400}{1800} = 3 \]

### Step 5: Finding the Third Wavelength
Now that we have \( n = 3 \), we can find the third wavelength \( \lambda_3 \) that will also be missing:
Using the condition for the third wavelength:
\[ 2 \mu t = (n + 2) \cdot \lambda_3 \]
Substituting \( n = 3 \):
\[ 2 \mu t = 5 \cdot \lambda_3 \]

From the first equation (for \( n = 3 \)):
\[ 2 \mu t = 3 \cdot 7200 \]
Setting the two equations equal gives:
\[ 3 \cdot 7200 = 5 \cdot \lambda_3 \]
Solving for \( \lambda_3 \):
\[ \lambda_3 = \frac{3 \cdot 7200}{5} = \frac{21600}{5} = 4320 \, \text{Å} \]

### Conclusion
The other wavelength in the visible region that is missing in the reflected spectrum is **4320 Å**.

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