Light is incident on a glass plate of refractive index 3.0 such that angle of refraction is 60° .Dark band is observed corresponding to the wavelength of 6000 A .If the thickness of glass plate is 1.2x`10^(-3)` mm. Calculate the order of the interference band for reflected system.

Answer

To solve the problem step by step, we will use the formula for dark bands in the reflected light system, which is given by:

\[ 2 \mu t \cos r = n \lambda \]

Where:
- \( \mu \) = refractive index of the glass plate
- \( t \) = thickness of the glass plate
- \( r \) = angle of refraction
- \( n \) = order of the interference band
- \( \lambda \) = wavelength of light

### Step 1: Gather the Given Data
- Refractive index, \( \mu = 3.0 \)
- Angle of refraction, \( r = 60^\circ \)
- Wavelength, \( \lambda = 6000 \) Å (angstrom)
- Thickness of glass plate, \( t = 1.2 \times 10^{-3} \) mm

### Step 2: Convert Units
1. Convert the wavelength from angstroms to meters:
\[ \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

2. Convert the thickness from millimeters to meters:
\[ t = 1.2 \times 10^{-3} \, \text{mm} = 1.2 \times 10^{-3} \times 10^{-3} \, \text{m} = 1.2 \times 10^{-6} \, \text{m} \]

### Step 3: Calculate \( \cos r \)
Calculate \( \cos 60^\circ \):
\[ \cos 60^\circ = \frac{1}{2} \]

### Step 4: Substitute Values into the Formula
Substituting the values into the formula:
\[ 2 \mu t \cos r = n \lambda \]
\[ 2 \times 3.0 \times (1.2 \times 10^{-6}) \times \left(\frac{1}{2}\right) = n \times (6 \times 10^{-7}) \]

### Step 5: Simplify the Left Side
Calculating the left side:
\[ 2 \times 3.0 \times (1.2 \times 10^{-6}) \times \frac{1}{2} = 3.0 \times (1.2 \times 10^{-6}) = 3.6 \times 10^{-6} \]

### Step 6: Solve for \( n \)
Now, we can solve for \( n \):
\[ 3.6 \times 10^{-6} = n \times (6 \times 10^{-7}) \]
\[ n = \frac{3.6 \times 10^{-6}}{6 \times 10^{-7}} = 6 \]

### Final Answer
The order of the interference band for the reflected system is:
\[ \boxed{6} \]