A glass plate has a thicknes t and refractive index `mu`. The angle of incidence of a ray from air into the plate is equal to the critical angle for glass air intrerface. The lateral shift (perpendicular distance between incident ray and emergent ray) of ray is given by

Options

A. `t(1-1/(sqrt(mu^(2)+1)))`
B. `mu(t-1/(sqrt(mu^(2)+1)))`
C. `t/(mu)(1-1/(mu^(2)+1))`
D. `(t-(mu)/(sqrt(mu^(2)-1)))`

Answer

To solve the problem of finding the lateral shift of a ray passing through a glass plate of thickness \( T \) and refractive index \( \mu \), we can follow these steps:

### Step 1: Understanding the Critical Angle
The critical angle \( \theta_C \) for the glass-air interface can be defined using Snell's law. Since the ray is coming from air (refractive index \( \mu_{\text{air}} = 1 \)) into glass (refractive index \( \mu_{\text{glass}} = \mu \)), we have:
\[ \sin \theta_C = \frac{\mu_{\text{air}}}{\mu_{\text{glass}}} = \frac{1}{\mu} \]

### Step 2: Drawing the Ray Diagram
Draw a diagram of the glass plate. The incident ray from air strikes the glass plate at the critical angle \( \theta_C \). The ray will refract into the glass and then emerge back into the air, making the emergent ray parallel to the incident ray.

### Step 3: Identifying Angles
When the ray enters the glass, it makes an angle \( \theta_C \) with the normal. The angle of refraction \( \phi \) can be found using Snell's law:
\[ \sin \theta_C = \mu \sin \phi \]
From this, we can express \( \sin \phi \):
\[ \sin \phi = \frac{\sin \theta_C}{\mu} = \frac{1/\mu}{\mu} = \frac{1}{\mu^2} \]

### Step 4: Finding the Distance Traveled in the Glass
The distance \( D \) that the ray travels in the glass can be expressed in terms of the thickness \( T \) and the angle \( \phi \):
\[ D = \frac{T}{\cos \phi} \]

### Step 5: Calculating the Lateral Shift
The lateral shift \( x_0 \) can be calculated using the formula:
\[ x_0 = D \sin(\theta_C - \phi) \]
Substituting \( D \) into this equation gives:
\[ x_0 = \frac{T}{\cos \phi} \sin(\theta_C - \phi) \]

### Step 6: Simplifying the Expression
Using the trigonometric identity for \( \sin(a - b) \):
\[ \sin(\theta_C - \phi) = \sin \theta_C \cos \phi - \cos \theta_C \sin \phi \]
Substituting values:
\[ x_0 = \frac{T}{\cos \phi} \left( \sin \theta_C \cos \phi - \cos \theta_C \sin \phi \right) \]
This simplifies to:
\[ x_0 = T \left( \sin \theta_C - \frac{\cos \theta_C \sin \phi}{\cos \phi} \right) \]

### Step 7: Final Calculation
Substituting \( \sin \theta_C = \frac{1}{\mu} \) and using the relationships for \( \sin \phi \) and \( \cos \theta_C \):
\[ x_0 = T \left( \frac{1}{\mu} - \tan \phi \right) \]
After substituting and simplifying further, we arrive at the final expression for the lateral shift.

### Final Result
The final expression for the lateral shift \( x_0 \) is:
\[ x_0 = \frac{T}{\mu} \sqrt{\mu^2 - 1} \]