If the plane surface of a plano-convex lens of radius of curvature R and refractive index `mu` is silvered, then its focal length would be

Options

A. `f = R/ 2(mu +1)`
B. `f = 2/ R(mu +1)`
C. `f = 2R/ (mu +1)`
D. `f = R/ 2(mu -1)`

Answer

To find the focal length of a plano-convex lens when its plane surface is silvered, we can follow these steps:

### Step 1: Understand the Lens Maker's Formula
The lens maker's formula for a lens in air is given by:

\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]

Where:
- \(\mu_2\) is the refractive index of the lens (which is \(\mu\)),
- \(\mu_1\) is the refractive index of air (which is 1),
- \(V\) is the image distance,
- \(U\) is the object distance,
- \(R\) is the radius of curvature of the lens.

### Step 2: Set Up the Equation for the Plano-Convex Lens
For a plano-convex lens, we consider a parallel beam of light incident on the convex surface. The object is at infinity, so \(U = -\infty\). Substituting these values into the lens maker's formula gives:

\[ \frac{\mu}{V_1} - \frac{1}{-\infty} = \frac{\mu - 1}{R} \]

Since \(\frac{1}{-\infty} = 0\), we simplify this to:

\[ \frac{\mu}{V_1} = \frac{\mu - 1}{R} \]

### Step 3: Solve for \(V_1\)
Rearranging the equation to solve for \(V_1\):

\[ V_1 = \frac{\mu R}{\mu - 1} \]

### Step 4: Image Reflection from the Silvered Surface
The image formed at \(V_1\) acts as an object for the reflection from the silvered plane surface. The distance of this image from the lens is \(V_1\), and it is now treated as an object for the second part of the lens.

### Step 5: Set Up the Equation for the Reflected Image
Using the lens maker's formula again, but now with the object distance as \(V_1\) and treating the silvered surface as a mirror (where the radius of curvature \(R\) is negative):

\[ \frac{1}{V_2} - \frac{\mu}{V_1} = \frac{1 - \mu}{-R} \]

Substituting \(V_1\) into this equation gives:

\[ \frac{1}{V_2} - \frac{\mu}{\frac{\mu R}{\mu - 1}} = \frac{1 - \mu}{-R} \]

### Step 6: Simplify the Equation
This simplifies to:

\[ \frac{1}{V_2} - \frac{\mu - 1}{R} = \frac{1 - \mu}{-R} \]

Combining the terms gives:

\[ \frac{1}{V_2} = \frac{2\mu - 1}{R} \]

### Step 7: Find the Focal Length
Since the power of the lens is defined as \(P = \frac{1}{f}\), where \(f\) is the focal length, and we have \(V_2\) as the image distance:

\[ \frac{1}{f} = \frac{1}{V_2} \]

Thus, we find:

\[ f = \frac{R}{2(\mu - 1)} \]

### Final Answer
The focal length of the silvered plano-convex lens is:

\[ f = \frac{R}{2(\mu - 1)} \]