Light of wavelength 3000A is incident on a thin glass plate of refractive index 0.5 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 find the thickness of a thin glass plate that will cause the reflected light to appear dark. This can be determined using the formula for destructive interference in thin films.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Wavelength of light, \( \lambda = 3000 \) Å (angstroms)
- Refractive index of glass, \( \mu = 0.5 \)
- Angle of refraction into the plate, \( r = 60^\circ \)

2. **Convert Wavelength to Meters:**
- Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \), we convert the wavelength:
\[ \lambda = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} = 3 \times 10^{-7} \, \text{m} \]

3. **Use the Formula for Thickness:**
- The condition for destructive interference (dark appearance) in reflected light is given by:
\[ 2 \mu t \cos r = n \lambda \]
where \( n \) is the order of the dark band. For the first dark band, we take \( n = 1 \).

4. **Rearranging the Formula for Thickness \( t \):**
- Rearranging gives:
\[ t = \frac{n \lambda}{2 \mu \cos r} \]

5. **Calculate \( \cos r \):**
- For \( r = 60^\circ \):
\[ \cos 60^\circ = \frac{1}{2} \]

6. **Substituting the Values into the Thickness Formula:**
- Now substituting \( n = 1 \), \( \lambda = 3 \times 10^{-7} \, \text{m} \), \( \mu = 0.5 \), and \( \cos 60^\circ = \frac{1}{2} \):
\[ t = \frac{1 \times 3 \times 10^{-7}}{2 \times 0.5 \times \frac{1}{2}} \]

7. **Simplifying the Expression:**
- The denominator simplifies to:
\[ 2 \times 0.5 \times \frac{1}{2} = 0.5 \]
- Thus, we have:
\[ t = \frac{3 \times 10^{-7}}{0.5} = 6 \times 10^{-7} \, \text{m} \]

8. **Final Answer:**
- The thickness of the plate that will make it appear dark by reflection is:
\[ t = 6 \times 10^{-7} \, \text{m} \, \text{or} \, 600 \, \text{nm} \]