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. `3/2`
B. `4/3`
C. `5/3`
D. 2

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the focal length of the lens
Given that the power \( P \) of the lens is \( +5.0D \), we can use the formula for power:
\[ P = \frac{1}{f} \]
where \( f \) is the focal length in meters. Rearranging gives:
\[ f = \frac{1}{P} = \frac{1}{5} \text{ meters} = 0.2 \text{ meters} = 20 \text{ cm} \]

### Step 2: Apply the lens maker's formula
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For the glass lens, substituting \( f = 20 \text{ cm} \) and \( \mu_g = 1.5 \):
\[ \frac{1}{20} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{1}{20} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Let’s denote this as Equation (1).

### Step 3: Determine the focal length in the liquid
When the lens is immersed in a liquid of refractive index \( \mu_1 \), it acts as a divergent lens with a focal length of \( -100 \text{ cm} \). Using the lens maker's formula again:
\[ \frac{1}{-100} = \left( \frac{1.5}{\mu_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{-1}{100} = \left( \frac{1.5}{\mu_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Let’s denote this as Equation (2).

### Step 4: Divide Equation (1) by Equation (2)
Dividing Equation (1) by Equation (2):
\[ \frac{\frac{1}{20}}{\frac{-1}{100}} = \frac{0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)}{\left( \frac{1.5}{\mu_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \]
This simplifies to:
\[ -5 = \frac{0.5}{\left( \frac{1.5}{\mu_1} - 1 \right)} \]
Cross-multiplying gives:
\[ -5 \left( \frac{1.5}{\mu_1} - 1 \right) = 0.5 \]
Expanding this:
\[ -7.5 + 5 = 0.5 \]
Thus:
\[ -2.5 = 0.5 \Rightarrow \frac{1.5}{\mu_1} = -2.5 \]

### Step 5: Solve for \( \mu_1 \)
Rearranging gives:
\[ \frac{1.5}{\mu_1} = -2.5 \Rightarrow \mu_1 = \frac{1.5}{-2.5} = \frac{5}{3} \]

### Final Answer
The refractive index \( \mu_1 \) of the liquid is:
\[ \mu_1 = \frac{5}{3} \approx 1.67 \]