A circular beam of light (diameter d) falls on a plane surface of a liquid. The angle of incidence is `45^(@)` and refractive index of the liquid is `mu`. The diameter of the refracted beam is

Options

A. d
B. `(mu-1)d`
C. `(sqrt(2mu^(2)-1))/dd`
D. `(sqrt(mu^(2)-1))/mud`

Explanation

To solve the problem of finding the diameter of the refracted beam when a circular beam of light with diameter \(d\) falls on a plane surface of a liquid at an angle of incidence of \(45^\circ\), we can follow these steps:

### Step 1: Understand the Geometry
We have a circular beam of light with diameter \(d\) incident on the surface of a liquid. The angle of incidence \(I\) is given as \(45^\circ\). The refractive index of the liquid is \(\mu\).

### Step 2: Apply Snell's Law
Using Snell's Law at the interface of air and liquid:
\[ \mu_{\text{air}} \sin I = \mu_{\text{liquid}} \sin R \]
Since \(\mu_{\text{air}} = 1\) and \(I = 45^\circ\), we have:
\[ \sin 45^\circ = \mu \sin R \]
This simplifies to:
\[ \frac{1}{\sqrt{2}} = \mu \sin R \]
Thus,
\[ \sin R = \frac{1}{\mu \sqrt{2}} \]

### Step 3: Calculate \(\cos R\)
Using the identity \(\cos^2 R + \sin^2 R = 1\):
\[ \cos^2 R = 1 - \sin^2 R = 1 - \left(\frac{1}{\mu \sqrt{2}}\right)^2 \]
This gives:
\[ \cos R = \sqrt{1 - \frac{1}{2\mu^2}} = \sqrt{\frac{2\mu^2 - 1}{2\mu^2}} \]

### Step 4: Relate the Diameters Before and After Refraction
From the geometry of the situation, we can relate the diameters before and after refraction using the triangles formed:
1. For the incident beam:
\[ AC = \frac{d}{\cos I} = \frac{d}{\cos 45^\circ} = \frac{d}{\frac{1}{\sqrt{2}}} = d \sqrt{2} \]
2. For the refracted beam:
\[ AC = \frac{D'}{\cos R} \]

Setting these equal gives:
\[ d \sqrt{2} = \frac{D'}{\cos R} \]
Thus,
\[ D' = d \sqrt{2} \cos R \]

### Step 5: Substitute \(\cos R\)
Substituting the expression for \(\cos R\):
\[ D' = d \sqrt{2} \cdot \sqrt{\frac{2\mu^2 - 1}{2\mu^2}} = d \cdot \sqrt{2} \cdot \frac{\sqrt{2\mu^2 - 1}}{\mu} \]

### Step 6: Final Expression for Diameter of Refracted Beam
Thus, the diameter of the refracted beam \(D'\) is:
\[ D' = \frac{d \sqrt{2(2\mu^2 - 1)}}{\mu} \]

### Final Answer
The diameter of the refracted beam is:
\[ D' = \frac{d}{\mu} \sqrt{2\mu^2 - 1} \]