A thin film of soap solution `(mu_(s)=1.4)` lies on the top of a glass plate `(mu_(g)=1.5)`. When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420 and 630 nm. The minimum thickness of the soap solution is

Options

A. 420 nm
B. 450 nm
C. 630 nm
D. 1260 nm

Answer

To find the minimum thickness of the soap solution, we can follow these steps:

### Step 1: Understand the setup
We have a thin film of soap solution (refractive index \( \mu_s = 1.4 \)) on a glass plate (refractive index \( \mu_g = 1.5 \)). When light is incident almost normally, reflections occur at both the soap-air interface and the soap-glass interface.

### Step 2: Phase change upon reflection
When light reflects off a medium with a higher refractive index, it undergoes a phase change of \( \pi \) (or half a wavelength). Here, the light reflecting from the soap-air interface (from lower to higher refractive index) will have a phase change of \( \pi \). The light reflecting from the soap-glass interface (from higher to lower refractive index) will also have a phase change of \( \pi \). Therefore, the total phase change is \( 2\pi \).

### Step 3: Condition for maxima
For constructive interference (maxima), the condition is given by:
\[ 2\mu t = n \lambda \]
where \( t \) is the thickness of the film, \( n \) is the order of the maximum, and \( \lambda \) is the wavelength of light in the medium.

### Step 4: Relate the two wavelengths
Given that adjacent maxima are observed at wavelengths \( \lambda_1 = 420 \, \text{nm} \) and \( \lambda_2 = 630 \, \text{nm} \), we can set up the relationship:
\[ n \lambda_1 = (n-1) \lambda_2 \]
Substituting the values:
\[ n \cdot 420 = (n - 1) \cdot 630 \]

### Step 5: Solve for \( n \)
Expanding and rearranging the equation:
\[ 420n = 630n - 630 \]
\[ 630 = 630n - 420n \]
\[ 630 = 210n \]
\[ n = \frac{630}{210} = 3 \]

### Step 6: Calculate the thickness
Now, substituting \( n = 3 \) into the condition for maxima using \( \lambda = 420 \, \text{nm} \):
\[ 2\mu t = n \lambda \]
\[ 2 \cdot 1.4 \cdot t = 3 \cdot 420 \]
\[ 2.8t = 1260 \]
\[ t = \frac{1260}{2.8} = 450 \, \text{nm} \]

### Final Answer
The minimum thickness of the soap solution is \( 450 \, \text{nm} \).

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