A beam of light falls on a glass plate `(mu=3//2)` of thickness 6.0 cm at an angle of `60^(@)`. Find the deflection of the beam on passing through the plate. (in cm).

Answer

To find the deflection of a beam of light passing through a glass plate, we can use the formula for deflection:

\[ D = d \sin \theta \left(1 - \sqrt{1 - \frac{\sin^2 \theta}{\mu^2}}\right) \]

where:
- \(D\) is the deflection,
- \(d\) is the thickness of the glass plate,
- \(\theta\) is the angle of incidence,
- \(\mu\) is the refractive index of the glass.

### Step 1: Identify the given values
- Thickness of the glass plate, \(d = 6.0 \, \text{cm}\)
- Angle of incidence, \(\theta = 60^\circ\)
- Refractive index of glass, \(\mu = \frac{3}{2} = 1.5\)

### Step 2: Calculate \(\sin \theta\)
Using the angle \(\theta = 60^\circ\):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Step 3: Substitute the values into the deflection formula
Substituting the values into the deflection formula:
\[ D = 6.0 \cdot \frac{\sqrt{3}}{2} \left(1 - \sqrt{1 - \frac{\left(\frac{\sqrt{3}}{2}\right)^2}{(1.5)^2}}\right) \]

### Step 4: Calculate \(\frac{\sin^2 \theta}{\mu^2}\)
Calculating \(\frac{\sin^2 60^\circ}{\mu^2}\):
\[ \frac{\sin^2 60^\circ}{\mu^2} = \frac{\left(\frac{\sqrt{3}}{2}\right)^2}{(1.5)^2} = \frac{\frac{3}{4}}{\frac{9}{4}} = \frac{3}{9} = \frac{1}{3} \]

### Step 5: Substitute this value back into the formula
Now substituting back:
\[ D = 6.0 \cdot \frac{\sqrt{3}}{2} \left(1 - \sqrt{1 - \frac{1}{3}}\right) \]

### Step 6: Simplify the expression inside the square root
Calculating \(1 - \frac{1}{3}\):
\[ 1 - \frac{1}{3} = \frac{2}{3} \]
Thus,
\[ \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} \]

### Step 7: Substitute this back into the deflection formula
Substituting this back:
\[ D = 6.0 \cdot \frac{\sqrt{3}}{2} \left(1 - \frac{\sqrt{2}}{\sqrt{3}}\right) \]

### Step 8: Calculate the final value
Calculating \(1 - \frac{\sqrt{2}}{\sqrt{3}}\):
\[ D = 6.0 \cdot \frac{\sqrt{3}}{2} \cdot \left(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}}\right) \]
\[ D = 6.0 \cdot \frac{1}{2} \cdot (\sqrt{3} - \sqrt{2}) = 3.0(\sqrt{3} - \sqrt{2}) \]

### Step 9: Numerical approximation
Using approximate values:
\(\sqrt{3} \approx 1.732\) and \(\sqrt{2} \approx 1.414\):
\[ D \approx 3.0(1.732 - 1.414) \approx 3.0(0.318) \approx 0.954 \, \text{cm} \]

Thus, the deflection of the beam on passing through the plate is approximately:
\[ D \approx 0.954 \, \text{cm} \]