When a lens of power `P` (in air) made of material of refractive index `mu` is immersed in liquid of refractive index `mu_(0)` Then the power of lens is:

Options

A. `(mu-1)/(mu-mu_(0))P`
B. `(mu-mu_(0))/(mu-1)P`
C. `(mu-mu_(0))/(mu-1)`. `(P)/(mu_(0))`
D. none of these

Explanation

To solve the problem of finding the power of a lens when it is immersed in a liquid, 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:
\[ \frac{1}{F} = \left(\mu - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Where:
- \( F \) is the focal length of the lens.
- \( \mu \) is the refractive index of the lens material.
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 2: Calculate the Power of the Lens in Air
The power \( P \) of the lens is defined as:
\[ P = \frac{1}{F} \]
When the lens is in air (where the refractive index of air is approximately 1), we can express the power as:
\[ P = \left(\mu - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

### Step 3: Calculate the Power of the Lens in Liquid
When the lens is immersed in a liquid with a refractive index \( \mu_0 \), the lens maker's formula modifies to:
\[ \frac{1}{F'} = \left(\frac{\mu}{\mu_0} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Here, \( F' \) is the new focal length of the lens in the liquid.

### Step 4: Express the New Power of the Lens
The new power \( P' \) of the lens in the liquid can be expressed as:
\[ P' = \frac{1}{F'} = \left(\frac{\mu}{\mu_0} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

### Step 5: Relate the Two Powers
To relate the two powers \( P \) and \( P' \), we can divide the equations for \( P' \) and \( P \):
\[ \frac{P'}{P} = \frac{\left(\frac{\mu}{\mu_0} - 1\right)}{\left(\mu - 1\right)} \]
Rearranging gives:
\[ P' = P \cdot \frac{\left(\frac{\mu}{\mu_0} - 1\right)}{\left(\mu - 1\right)} \]

### Step 6: Substitute and Simplify
Substituting \( \frac{\mu}{\mu_0} - 1 = \frac{\mu - \mu_0}{\mu_0} \):
\[ P' = P \cdot \frac{\left(\frac{\mu - \mu_0}{\mu_0}\right)}{\left(\mu - 1\right)} \]
This simplifies to:
\[ P' = P \cdot \frac{\mu - \mu_0}{\mu - 1} \cdot \frac{1}{\mu_0} \]

### Final Result
Thus, the power of the lens when immersed in a liquid is given by:
\[ P' = \frac{(\mu - \mu_0)}{(\mu - 1) \cdot \mu_0} \cdot P \]