A convex lens (refractive index `mu` = 1.5) has a power P. If it is immersed in a liquid (`mu` = 4/3), then its power will become/remain

Options

A. P
B. `P/2`
C. `P/4`
D. 4P

Answer

To solve the problem, we need to determine how the power of a convex lens changes when it is immersed in a liquid with a different refractive index. Here’s a step-by-step solution:

### Step 1: Understand the concept of lens power
The power \( P \) of a lens is defined as the inverse of its focal length \( f \):
\[ P = \frac{1}{f} \]
The focal length of a lens can be derived from the lens maker's formula.

### Step 2: Write the lens maker's formula
The lens maker's formula is given by:
\[ \frac{1}{f} = \left( \mu_{\text{relative}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( \mu_{\text{relative}} \) is the relative refractive index of the lens material with respect to the surrounding medium.

### Step 3: Calculate the relative refractive index in air
For the lens in air, the refractive index of the lens \( \mu_{\text{lens}} = 1.5 \) and the refractive index of air \( \mu_{\text{air}} = 1 \):
\[ \mu_{\text{relative}} = \frac{\mu_{\text{lens}}}{\mu_{\text{air}}} = \frac{1.5}{1} = 1.5 \]

### Step 4: Calculate the power in air
Using the lens maker's formula, the power \( P \) in air can be expressed as:
\[ P = \left( 1.5 - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 5: Calculate the relative refractive index in liquid
Now, when the lens is immersed in a liquid with a refractive index \( \mu_{\text{liquid}} = \frac{4}{3} \):
\[ \mu_{\text{relative}} = \frac{\mu_{\text{lens}}}{\mu_{\text{liquid}}} = \frac{1.5}{\frac{4}{3}} = \frac{1.5 \times 3}{4} = \frac{4.5}{4} = 1.125 \]

### Step 6: Calculate the new power in liquid
Using the lens maker's formula again, the new power \( P' \) in the liquid becomes:
\[ P' = \left( 1.125 - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.125 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 7: Relate the powers in air and liquid
Now we can relate the two powers:
\[ \frac{P'}{P} = \frac{0.125}{0.5} = \frac{1}{4} \]
Thus, we find:
\[ P' = \frac{1}{4} P \]

### Conclusion
When the convex lens is immersed in the liquid, its power becomes one-fourth of its original power.