Question

Consider a glass slab which is silvered at one side and the other side is transparent, given the refractive index of the class slab to be 1.5 if a Ray of light is incident at an angle at 45 degree on the transparent side the deviation of the ray of life from its initial path when comes out of the slab is

• A. 120 degree
• B. 90 degree
• C. 45 degree
• D. 180 degree

Answer 1

The Correct Option is B

A ray of light enters the transparent side of a glass slab at an angle of incidence \(\theta\). Due to the refractive index \(\mu = 1.5\), the ray undergoes refraction and bends towards the normal inside the slab.

It then reaches the silvered surface (back side), where it undergoes reflection (as if hitting a mirror). The reflected ray travels back through the slab and finally emerges from the initial side into the air, undergoing refraction again.

In this setup, due to the combination of refraction and reflection, the emergent ray deviates by \(90^\circ\) from its original direction. This is a special case where the light path effectively makes a right-angle turn due to the geometry and nature of the slab.

Thus, the deviation is: \(\boxed{90^\circ}\)

Answer 2

Let the angle of incidence on the transparent side be \( i_1 = 45^\circ \).  
The refractive index of the glass slab is \( n_2 = 1.5 \), and the refractive index of air is \( n_1 \approx 1 \).

The path of the light ray involves three stages:

1. Refraction as it enters the glass slab from air.
2. Reflection from the silvered surface at the back.
3. Refraction as it exits the glass slab back into the air.

 

Step 1: Refraction at the first surface (Air to Glass)
Using Snell's Law: \( n_1 \sin i_1 = n_2 \sin r_1 \), where \( r_1 \) is the angle of refraction inside the glass. \[ 1 \times \sin 45^\circ = 1.5 \times \sin r_1 \] \[ \frac{1}{\sqrt{2}} = 1.5 \sin r_1 \] \[ \sin r_1 = \frac{1}{1.5 \sqrt{2}} = \frac{1}{(3/2)\sqrt{2}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3} \] So, the angle of refraction inside the glass is \( r_1 = \arcsin\left(\frac{\sqrt{2}}{3}\right) \).

Step 2: Reflection at the silvered surface
The refracted ray travels through the glass and strikes the silvered surface. Assuming the slab faces are parallel, the angle of incidence on the silvered surface is \( i_2 = r_1 \). 
According to the law of reflection, the angle of reflection \( r_2 \) is equal to the angle of incidence: \( r_2 = i_2 = r_1 \).

Step 3: Refraction at the first surface (Glass to Air)
The reflected ray travels back through the glass and strikes the transparent surface from the inside. Since the faces are parallel, the angle of incidence on this surface is \( i_3 = r_2 = r_1 \). 
Let the angle of emergence into the air be \( r_3 \). Using Snell's Law again: \[ n_2 \sin i_3 = n_1 \sin r_3 \] \[ 1.5 \sin r_1 = 1 \times \sin r_3 \] From Step 1, we know \( 1.5 \sin r_1 = 1 \times \sin 45^\circ \). Substituting this into the equation: \[ \sin 45^\circ = \sin r_3 \] Therefore, the angle of emergence is \( \mathbf{r_3 = 45^\circ} \).

Step 4: Calculating the Total Deviation
The incident ray strikes the surface at an angle \( i_1 = 45^\circ \) with the normal. 
The final emergent ray leaves the surface at an angle \( r_3 = 45^\circ \) with the normal. 
Crucially, because the ray has undergone reflection, the emergent ray is on the same side of the normal as the incident ray.

The deviation (\( \delta \)) is the angle between the initial direction of the incident ray and the final direction of the emergent ray. Let's consider the initial ray's path extended straight. The final ray emerges at an angle \( r_3 = i_1 = 45^\circ \) to the normal, but directed back towards the side of incidence.

Imagine the incident ray and the emergent ray. Both make an angle of \( 45^\circ \) with the normal. If the emergent ray was on the opposite side of the normal (like transmission through a non-silvered slab), the deviation would be 0 (rays are parallel). Since it emerges on the same side, we can think of the overall process as equivalent to a reflection.

The final emergent ray is parallel to a ray that would be reflected if the front surface itself were a mirror with the incident angle \( i_1 \). The deviation caused by reflection from a plane surface is given by: \[ \delta = 180^\circ - 2 \times (\text{angle of incidence}) \] In this scenario, the effective angle of incidence determining the overall deviation is the initial angle \( i_1 \). \[ \mathbf{\delta = 180^\circ - 2 i_1} \] Substituting \( i_1 = 45^\circ \): \[ \delta = 180^\circ - 2(45^\circ) \] \[ \delta = 180^\circ - 90^\circ \] \[ \mathbf{\delta = 90^\circ} \]

Note that the refractive index of the glass slab (\( n=1.5 \)) influences the path of the ray inside the slab and causes a lateral displacement, but it does not affect the final angle of deviation in this specific configuration (parallel slab, reflection from a parallel back surface).

Therefore, the deviation of the ray of light from its initial path is 90 degrees.