There is layer of medium of variable refractive index `mu = sqrt(2) - (1)/(sqrt(2)) y` (where `0 le t lt 1//2`) sandwitched between the layer of glass and air A beam of light travelling in air at an angle `45^(@)` has a width `Delta omega`. When the beam enters the layer of glass its width becomes : `(mu_("glass") = sqrt(2))`

Options

A. a. `sqrt((3)/(2)) Delta omega`
B. b. `sqrt((2)/(3)) Delta omega`
C. c. `sqrt((1)/(3)) Delta omega`
D. d. `Delta omega`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the setup
We have a layer of medium with a variable refractive index given by:
\[ \mu = \sqrt{2} - \frac{1}{\sqrt{2}} y \]
This layer is sandwiched between glass (with refractive index \(\mu_{\text{glass}} = \sqrt{2}\)) and air (with refractive index \(\mu_{\text{air}} = 1\)). A beam of light enters the medium from air at an angle of \(45^\circ\).

### Step 2: Apply Snell's Law at the air-glass interface
Using Snell's Law:
\[ \mu_1 \sin i = \mu_2 \sin r \]
where:
- \(\mu_1 = 1\) (air)
- \(i = 45^\circ\)
- \(\mu_2 = \sqrt{2}\) (glass)
- \(r\) is the angle of refraction in glass.

Substituting the values:
\[ 1 \cdot \sin(45^\circ) = \sqrt{2} \cdot \sin(r) \]
\[ \frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin(r) \]
\[ \sin(r) = \frac{1}{2} \]
Thus, \(r = 30^\circ\).

### Step 3: Determine the angle of incidence in the variable medium
Since the beam of light enters the variable refractive index medium, we need to find the angle of incidence in this medium. The angle of incidence can be calculated as:
\[ \theta = 90^\circ - r = 90^\circ - 30^\circ = 60^\circ \]

### Step 4: Calculate the width of the beam in the variable medium
Let \(h\) be the thickness of the variable medium. The width of the beam in the variable medium can be expressed as:
\[ \Delta \omega' = h \cdot \tan(\theta) \]
where \(\theta = 60^\circ\):
\[ \Delta \omega' = h \cdot \tan(60^\circ) = h \cdot \sqrt{3} \]

### Step 5: Relate the width in air to the width in glass
In air, the width of the beam is given by:
\[ \Delta \omega = h \cdot \tan(45^\circ) = h \]

### Step 6: Find the relationship between the widths
Now, we can relate the widths:
\[ \Delta \omega' = \sqrt{3} \cdot \Delta \omega \]

### Conclusion
Thus, the width of the beam when it enters the layer of glass becomes:
\[ \Delta \omega' = \frac{\sqrt{3}}{2} \Delta \omega \]