A light ray is incident at an angle of `45^@` with the normal to a `sqrt2` cm thick plate` (mu=2.0)`. Find the shift in the path of the light as it emerges out from the plate.

Explanation

To find the shift in the path of light as it emerges from a plate, we can follow these steps:

### Step 1: Identify the Given Values
- Angle of incidence (I) = 45°
- Thickness of the plate (T) = √2 cm
- Refractive index of the plate (μ) = 2.0
- Refractive index of air (n1) = 1.0

### Step 2: Use Snell's Law to Find the Angle of Refraction (R)
Using Snell's Law:
\[ n_1 \sin I = n_2 \sin R \]
Substituting the known values:
\[ 1 \cdot \sin(45°) = 2 \cdot \sin R \]
Since \(\sin(45°) = \frac{1}{\sqrt{2}}\), we have:
\[ \frac{1}{\sqrt{2}} = 2 \cdot \sin R \]
Solving for \(\sin R\):
\[ \sin R = \frac{1}{2\sqrt{2}} = \frac{1}{2\sqrt{2}} \approx 0.3536 \]
Now, we can find the angle R:
\[ R \approx 20.7° \]

### Step 3: Calculate the Lateral Displacement (Δ)
The formula for lateral displacement (Δ) is given by:
\[ \Delta = T \cdot \frac{\sin(I - R)}{\cos R} \]
Substituting the values we have:
- \(T = \sqrt{2} \text{ cm}\)
- \(I = 45°\)
- \(R \approx 20.7°\)

Calculating \(I - R\):
\[ I - R = 45° - 20.7° = 24.3° \]

Now substituting into the formula:
\[ \Delta = \sqrt{2} \cdot \frac{\sin(24.3°)}{\cos(20.7°)} \]

### Step 4: Calculate \(\sin(24.3°)\) and \(\cos(20.7°)\)
Using a calculator:
- \(\sin(24.3°) \approx 0.409\)
- \(\cos(20.7°) \approx 0.935\)

### Step 5: Substitute and Solve
Now substituting these values into the equation:
\[ \Delta = \sqrt{2} \cdot \frac{0.409}{0.935} \]
Calculating:
\[ \Delta \approx 1.414 \cdot 0.438 \approx 0.620 \text{ cm} \]

### Final Result
Thus, the shift in the path of light as it emerges from the plate is approximately:
\[ \Delta \approx 0.622 \text{ cm} \]