A ray of light is incident from air on a glass plate having thickness `sqrt3` cm and refractive index `sqrt2` . The angle of incidence of a ray is equal to the cirtical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is -------- × `10^(–2) cm.` (given `sin 15^@ = 0.26`)

Answer

To solve the problem, we need to calculate the lateral displacement of a ray of light passing through a glass plate. Let's break it down step by step.

### Step 1: Identify the critical angle
The critical angle \( C \) for the glass-air interface can be calculated using Snell's law. The refractive index of air is approximately 1, and the refractive index of glass is given as \( \sqrt{2} \).

Using Snell's law:
\[ \sin C = \frac{n_1}{n_2} = \frac{1}{\sqrt{2}} \]
Thus, the critical angle \( C \) is:
\[ C = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \]

### Step 2: Determine the angle of incidence
The problem states that the angle of incidence \( i \) is equal to the critical angle, which means:
\[ i = 45^\circ \]

### Step 3: Calculate the angle of refraction
Using Snell's law again at the air-glass interface:
\[ n_1 \sin i = n_2 \sin r \]
Substituting the values:
\[ 1 \cdot \sin(45^\circ) = \sqrt{2} \cdot \sin r \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin r \]
Rearranging gives:
\[ \sin r = \frac{1}{2} \implies r = 30^\circ \]

### Step 4: Calculate the lateral displacement
The formula for lateral displacement \( d \) is given by:
\[ d = t \cdot \frac{\sin(i - r)}{\cos r} \]
Where:
- \( t = \sqrt{3} \) cm (thickness of the glass plate),
- \( i = 45^\circ \),
- \( r = 30^\circ \).

Substituting these values:
\[ d = \sqrt{3} \cdot \frac{\sin(45^\circ - 30^\circ)}{\cos(30^\circ)} \]
Calculating \( \sin(15^\circ) \) and \( \cos(30^\circ) \):
\[ \sin(15^\circ) = 0.26 \quad \text{and} \quad \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Now substituting these values:
\[ d = \sqrt{3} \cdot \frac{0.26}{\frac{\sqrt{3}}{2}} = \sqrt{3} \cdot \frac{0.26 \cdot 2}{\sqrt{3}} = 2 \cdot 0.26 = 0.52 \text{ cm} \]

### Step 5: Convert to the required format
The final answer needs to be expressed in the format of \( \text{value} \times 10^{-2} \text{ cm} \):
\[ 0.52 \text{ cm} = 52 \times 10^{-2} \text{ cm} \]

### Final Answer
The lateral displacement of the ray when it passes through the plate is \( 52 \times 10^{-2} \) cm.

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