A thin convergent glass lens `(mu_g=1.5)` has a power of `+5.0D.` When this lens is immersed in a liquid of refractive index `mu_1,` it acts as a divergent lens of focal length `100 cm.` The value of `mu_1` is

Options

A. `4//3`
B. `5//3`
C. `5//4`
D. `6//5`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the given data
- The refractive index of glass, \( \mu_g = 1.5 \)
- The power of the lens, \( P = +5.0 \, D \)
- When immersed in a liquid of refractive index \( \mu_1 \), the lens acts as a divergent lens with a focal length of \( F = -100 \, cm \).

### Step 2: Calculate the focal length of the lens in air
The power \( P \) of a lens is related to its focal length \( F \) by the formula:
\[ P = \frac{1}{F} \quad \text{(in meters)} \]
Given that \( P = +5.0 \, D \), we can find the focal length \( F \):
\[ F = \frac{1}{P} = \frac{1}{5.0} = 0.2 \, m = 20 \, cm \]

### Step 3: Use the lens maker's formula for the lens in air
The lens maker's formula is given by:
\[ \frac{1}{F} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For our converging lens in air:
\[ \frac{1}{20} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ \frac{1}{20} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Let’s denote \( \frac{1}{R_1} - \frac{1}{R_2} \) as \( k \). Thus:
\[ \frac{1}{20} = 0.5k \implies k = \frac{1}{10} \]

### Step 4: Use the lens maker's formula for the lens in the liquid
When the lens is immersed in a liquid, it behaves as a divergent lens with a focal length of \( -100 \, cm \):
\[ \frac{1}{F} = \left(\frac{\mu_g}{\mu_1} - 1\right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting \( F = -100 \, cm \):
\[ \frac{1}{-100} = \left(\frac{1.5}{\mu_1} - 1\right) k \]
Substituting \( k = \frac{1}{10} \):
\[ \frac{1}{-100} = \left(\frac{1.5}{\mu_1} - 1\right) \frac{1}{10} \]
Multiplying both sides by \( -10 \):
\[ \frac{1}{10} = 1 - \frac{1.5}{\mu_1} \]
Rearranging gives:
\[ \frac{1.5}{\mu_1} = 1 - \frac{1}{10} = \frac{9}{10} \]
Thus:
\[ \mu_1 = \frac{1.5 \times 10}{9} = \frac{15}{9} = \frac{5}{3} \]

### Final Answer
The refractive index of the liquid \( \mu_1 \) is \( \frac{5}{3} \).

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