A light ray is incident on a transparent slab of refractive index `mu=sqrt(2)` at an angle of incidence `pi//4`. Find the ratio of the lateral displacement suffered by the light ray to the maximum value which it could have suffered.

Options

A. `(sqrt(3)-1)/(sqrt(6))`
B. `(sqrt(3)-2)/(sqrt(5))`
C. `(sqrt(1)-2)/(sqrt(5))`
D. `(sqrt(1)-2)/(sqrt(7))`

Answer

To solve the problem, we need to find the lateral displacement of a light ray passing through a transparent slab and then calculate the ratio of this displacement to its maximum possible value. Here’s the step-by-step solution:

### Step 1: Understand the parameters given
- Refractive index of the slab, \( \mu = \sqrt{2} \)
- Angle of incidence, \( i = \frac{\pi}{4} \) (which is 45 degrees)

### Step 2: Use Snell's Law to find the angle of refraction
According to Snell's Law:
\[ \mu = \frac{\sin i}{\sin r} \]
Substituting the values we have:
\[ \sqrt{2} = \frac{\sin\left(\frac{\pi}{4}\right)}{\sin r} \]
Since \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \):
\[ \sqrt{2} = \frac{\frac{\sqrt{2}}{2}}{\sin r} \]
Rearranging gives:
\[ \sin r = \frac{\frac{\sqrt{2}}{2}}{\sqrt{2}} = \frac{1}{2} \]
Thus, \( r = \frac{\pi}{6} \) (which is 30 degrees).

### Step 3: Calculate the lateral displacement \( x \)
The formula for lateral displacement \( x \) through a slab is given by:
\[ x = t \cdot \frac{\sin i - \sin r}{\cos r} \]
Where \( t \) is the thickness of the slab. We can express \( \sin i \) and \( \sin r \):
- \( \sin i = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
- \( \sin r = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)
- \( \cos r = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)

Substituting these values into the formula:
\[ x = t \cdot \frac{\frac{\sqrt{2}}{2} - \frac{1}{2}}{\frac{\sqrt{3}}{2}} = t \cdot \frac{\frac{\sqrt{2} - 1}{2}}{\frac{\sqrt{3}}{2}} = t \cdot \frac{\sqrt{2} - 1}{\sqrt{3}} \]

### Step 4: Find the maximum lateral displacement
The maximum lateral displacement occurs when the angle of incidence is such that the light ray is incident normally (i.e., \( i = 90^\circ \)). In this case, the maximum lateral displacement \( x_{max} \) can be calculated as:
\[ x_{max} = t \cdot \frac{\sin i}{\cos r} \]
Since \( \sin(90^\circ) = 1 \):
\[ x_{max} = t \cdot \frac{1}{\cos r} = t \cdot \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2t}{\sqrt{3}} \]

### Step 5: Calculate the ratio of lateral displacement to maximum displacement
Now we find the ratio:
\[ \text{Ratio} = \frac{x}{x_{max}} = \frac{t \cdot \frac{\sqrt{2} - 1}{\sqrt{3}}}{\frac{2t}{\sqrt{3}}} \]
Cancelling \( t \) and \( \sqrt{3} \):
\[ \text{Ratio} = \frac{\sqrt{2} - 1}{2} \]

### Final Answer
The ratio of the lateral displacement suffered by the light ray to the maximum value which it could have suffered is:
\[ \frac{\sqrt{2} - 1}{2} \]