The outer surface of a transparent glass slab of refractive index `mu_(S)=1.5` is coated by a thin layer of transparent medium of refractive index `mu_(C)=1.6`. Orange light of wavelength 6400Å falls normally on the coating. The reflected light at the upper surface and at the lower surface of the coat interfere destructively. If thickness of the coat is `5Kxx10^(-8)`m, calculate the minimum value of K.

Answer

To solve the problem, we need to find the minimum value of \( K \) given the conditions of destructive interference for light reflected from the upper and lower surfaces of a thin coating of refractive index \( \mu_C = 1.6 \) on a glass slab of refractive index \( \mu_S = 1.5 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a thin coating of refractive index \( \mu_C = 1.6 \) on a glass slab with refractive index \( \mu_S = 1.5 \).
- Orange light of wavelength \( \lambda = 6400 \, \text{Å} = 6400 \times 10^{-10} \, \text{m} \) is incident normally on the coating.
- The thickness of the coating is given as \( t = 5K \times 10^{-8} \, \text{m} \).

2. **Condition for Destructive Interference**:
- For destructive interference, the path difference between the light reflected from the upper surface and the lower surface of the coating must be an odd multiple of half the wavelength:
\[ \Delta x = 2t \mu_C - \frac{\lambda}{2} = \left(2n - 1\right) \frac{\lambda}{2} \]
- Here, \( n \) is an integer representing the order of interference.

3. **Calculating the Path Difference**:
- The path difference can be expressed as:
\[ \Delta x = 2 \mu_C t - \frac{\lambda}{2} \]
- Setting this equal to the condition for destructive interference:
\[ 2 \mu_C t - \frac{\lambda}{2} = \left(2n - 1\right) \frac{\lambda}{2} \]

4. **Rearranging the Equation**:
- Rearranging gives:
\[ 2 \mu_C t = \left(2n\right) \frac{\lambda}{2} \]
- Simplifying this, we have:
\[ \mu_C t = n \frac{\lambda}{2} \]

5. **Substituting Values**:
- Substituting \( \mu_C = 1.6 \) and \( \lambda = 6400 \times 10^{-10} \, \text{m} \):
\[ 1.6 t = n \frac{6400 \times 10^{-10}}{2} \]
- For the minimum value of \( K \), we take \( n = 1 \):
\[ 1.6 t = 1 \cdot \frac{6400 \times 10^{-10}}{2} \]
- This simplifies to:
\[ 1.6 t = 3200 \times 10^{-10} \]

6. **Solving for Thickness \( t \)**:
- Solving for \( t \):
\[ t = \frac{3200 \times 10^{-10}}{1.6} = 2000 \times 10^{-10} \, \text{m} = 2 \times 10^{-7} \, \text{m} \]

7. **Relating Thickness to \( K \)**:
- We know \( t = 5K \times 10^{-8} \):
\[ 2 \times 10^{-7} = 5K \times 10^{-8} \]
- Dividing both sides by \( 10^{-8} \):
\[ 20 = 5K \]
- Therefore, solving for \( K \):
\[ K = \frac{20}{5} = 4 \]

### Final Answer:
The minimum value of \( K \) is \( 4 \).