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 relationship between power and focal length
The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{F} \]
where \( F \) is the focal length of the lens.

### Step 2: Use the lens maker's formula
The lens maker's formula relates the focal length \( F \) of a lens to its refractive index and the radii of curvature of its surfaces:
\[ \frac{1}{F} = \mu_R \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( \mu_R \) is the relative refractive index of the lens with respect to the medium it is in.

### Step 3: Calculate the relative refractive index in air
In air, the refractive index of the lens \( \mu = 1.5 \) and the refractive index of air \( \mu_{air} = 1 \). Thus, the relative refractive index \( \mu_R \) when the lens is in air is:
\[ \mu_R = \frac{\mu_{lens}}{\mu_{air}} = \frac{1.5}{1} = 1.5 \]

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

### Step 5: Calculate the relative refractive index in liquid
When the lens is immersed in a liquid with a refractive index \( \mu_{liquid} = \frac{4}{3} \), the new relative refractive index \( \mu_R \) becomes:
\[ \mu_R = \frac{\mu_{lens}}{\mu_{liquid}} = \frac{1.5}{\frac{4}{3}} = \frac{1.5 \times 3}{4} = \frac{4.5}{4} = \frac{9}{8} \]

### Step 6: Calculate the new power in the liquid
Using the lens maker's formula again for the liquid:
\[ \frac{1}{F'} = \frac{9}{8} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Thus, the new power \( P' \) in the liquid is:
\[ P' = \frac{9}{8} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 7: Relate the powers in air and liquid
Now we can relate the two powers:
\[ \frac{P}{P'} = \frac{1.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)}{\frac{9}{8} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \]
This simplifies to:
\[ \frac{P}{P'} = \frac{1.5}{\frac{9}{8}} = \frac{1.5 \times 8}{9} = \frac{12}{9} = \frac{4}{3} \]
Thus, we can find \( P' \):
\[ P' = \frac{P \times 3}{4} \]

### Step 8: Conclusion
The new power \( P' \) when the lens is immersed in the liquid is:
\[ P' = \frac{P}{4} \]

### Final Answer
The power of the lens when immersed in the liquid becomes \( \frac{P}{4} \).

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