Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle `theta` . The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

Options

A. `(4pi d)/(lambda)(1-(1)/(n^(2))"sin"^(2)theta)^(1//2)+pi`
B. `(4pi d)/(lambda)(1-(1)/(n^(2))"sin"^(2)theta)^(1//2)`
C. `(4pi d)/(lambda)(1-(1)/(n^(2))"sin"^(2)theta)^(1//2)+(pi)/(2)`
D. `(4pi d)/(lambda)(1-(1)/(n^(2))"sin"^(2)theta)^(1//2)+2pi`

Answer

To find the phase difference between the ray reflected from the top surface of a glass slab and the ray reflected from the bottom surface, we can follow these steps:

### Step 1: Understand the setup
We have a ray of light incident from air (refractive index = 1) onto a glass slab (refractive index = n) at an angle θ. The glass slab has a width d.

**Hint:** Visualize the situation with a diagram showing the incident ray, reflected rays, and the normal line.

### Step 2: Identify the rays
When the light hits the top surface of the glass slab, part of it is reflected back (let's call this Ray 1), and part of it enters the glass slab. The ray that enters the slab will eventually reflect off the bottom surface of the slab (let's call this Ray 2) and then exit back into the air.

**Hint:** Remember that Ray 1 is reflected at the top surface, while Ray 2 is reflected at the bottom surface.

### Step 3: Calculate the path difference
The path difference (Δp) between Ray 1 and Ray 2 can be determined by considering the geometry of the rays. The ray that enters the slab travels a distance d and then reflects back, while the reflected ray from the top surface travels a shorter distance.

The effective path difference can be expressed as:
\[ \Delta p = 2d \cos r \]
where \( r \) is the angle of refraction at the glass-air interface.

**Hint:** Use trigonometric relationships to express \( \cos r \) in terms of θ and n.

### Step 4: Apply Snell's Law
Using Snell's Law, we have:
\[ \sin \theta = n \sin r \]
From this, we can derive:
\[ \sin r = \frac{\sin \theta}{n} \]
Using the identity \( \cos^2 r + \sin^2 r = 1 \), we can find \( \cos r \):
\[ \cos r = \sqrt{1 - \left(\frac{\sin \theta}{n}\right)^2} \]

**Hint:** Remember that \( \cos r \) can also be expressed in terms of \( n \) and \( \theta \).

### Step 5: Substitute into the path difference
Now substitute \( \cos r \) back into the expression for path difference:
\[ \Delta p = 2d \sqrt{1 - \left(\frac{\sin \theta}{n}\right)^2} \]

**Hint:** This expression gives you the total path difference that contributes to the phase difference.

### Step 6: Calculate the phase difference
The phase difference (Δφ) between the two rays is given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta p \]
Substituting for Δp:
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot 2d \sqrt{1 - \left(\frac{\sin \theta}{n}\right)^2} \]

### Step 7: Account for the phase change upon reflection
When light reflects off a medium with a higher refractive index (from air to glass), it undergoes a phase change of π (or 180 degrees). Since Ray 1 reflects off the top surface, it experiences this phase change. Ray 2 does not experience a phase change upon reflection at the bottom surface.

Thus, the total phase difference becomes:
\[ \Delta \phi = \frac{4\pi d}{\lambda} \sqrt{1 - \left(\frac{\sin \theta}{n}\right)^2} + \pi \]

### Final Result
The final expression for the phase difference between the two rays is:
\[ \Delta \phi = \frac{4\pi d}{\lambda} \sqrt{1 - \left(\frac{\sin \theta}{n}\right)^2} + \pi \]

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