Light of wavelength 12000A is incident on a thin glass plate of refractive index 1.5 such that angle of refraction into plate is 60° .calculate the thickness of plate which will make it appear dark by reflection?

Answer

To solve the problem, we need to calculate the thickness of a thin glass plate that will make it appear dark by reflection when light of a specific wavelength is incident on it. The key concept here is the interference of light due to a thin film.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Wavelength of light, \( \lambda = 12000 \, \text{Å} = 12000 \times 10^{-10} \, \text{m} = 12 \times 10^{-7} \, \text{m} \)
- Refractive index of glass, \( \mu = 1.5 \)
- Angle of refraction, \( r = 60^\circ \)

2. **Understand the Condition for Dark Fringes:**
- For destructive interference (dark fringe) in a thin film, the path difference must be an odd multiple of half wavelengths. The formula for the path difference is given by:
\[ 2\mu d \cos r = (n + \frac{1}{2}) \lambda \]
- For the first dark fringe, \( n = 0 \):
\[ 2\mu d \cos r = \frac{1}{2} \lambda \]

3. **Rearranging the Formula:**
- We need to find the thickness \( d \):
\[ d = \frac{n \lambda}{2\mu \cos r} \]
- For the first dark fringe, substitute \( n = 0 \):
\[ d = \frac{\lambda}{2\mu \cos r} \]

4. **Calculating \( \cos r \):**
- Given \( r = 60^\circ \):
\[ \cos 60^\circ = \frac{1}{2} \]

5. **Substituting Values into the Thickness Formula:**
- Now substitute \( \lambda \), \( \mu \), and \( \cos r \) into the equation:
\[ d = \frac{12 \times 10^{-7} \, \text{m}}{2 \times 1.5 \times \frac{1}{2}} \]
- Simplifying this:
\[ d = \frac{12 \times 10^{-7}}{1.5} \]

6. **Final Calculation:**
- Calculate \( d \):
\[ d = 8 \times 10^{-7} \, \text{m} \]

### Final Answer:
The thickness of the plate that will make it appear dark by reflection is \( d = 8 \times 10^{-7} \, \text{m} \) or \( 800 \, \text{nm} \).

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