A parallel beam of sodium light of wavelength `6000 Å` is incident on a thin glass plate of `mu=1.5`, such that the angle of refraction in the plate is `60^(@)`. The smallest thickness of the plate which will make it appear dark by reflected light is

Options

A. `40 Å`
B. `4 Å`
C. `400 Å`
D. `4000 Å`

Explanation

To find the smallest thickness of the glass plate that will make it appear dark by reflected light, we can follow these steps:

### Step 1: Understand the condition for dark fringes
For destructive interference (dark fringes) in reflected light, the path difference must satisfy the condition:
\[ \text{Path difference} = (2n + 1) \frac{\lambda}{2} \]
where \( n \) is an integer (0, 1, 2, ...), and \( \lambda \) is the wavelength of the light.

### Step 2: Identify the parameters
Given:
- Wavelength of sodium light, \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- Refractive index of the glass plate, \( \mu = 1.5 \)
- Angle of refraction in the plate, \( r = 60^\circ \)

### Step 3: Calculate the extra path length
When light travels through the glass plate, the extra path length due to reflection at the top and bottom surfaces of the plate can be expressed as:
\[ \text{Extra path length} = 2 \mu t \cos r \]

### Step 4: Set up the equation for dark fringe
For the smallest thickness that results in a dark fringe, we can set \( n = 0 \):
\[ 2 \mu t \cos r = (2n + 1) \frac{\lambda}{2} + \frac{\lambda}{2} \]
This simplifies to:
\[ 2 \mu t \cos r = \lambda \]

### Step 5: Substitute known values
Substituting the known values into the equation:
\[ 2 \times 1.5 \times t \times \cos(60^\circ) = 6 \times 10^{-7} \]
Since \( \cos(60^\circ) = \frac{1}{2} \), we have:
\[ 2 \times 1.5 \times t \times \frac{1}{2} = 6 \times 10^{-7} \]
This simplifies to:
\[ 1.5 t = 6 \times 10^{-7} \]

### Step 6: Solve for thickness \( t \)
Now, solving for \( t \):
\[ t = \frac{6 \times 10^{-7}}{1.5} = 4 \times 10^{-7} \, \text{m} = 4000 \, \text{Å} \]

### Final Answer
Thus, the smallest thickness of the plate which will make it appear dark by reflected light is:
\[ t = 4000 \, \text{Å} \]

---