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 will follow these steps:

### Step 1: Understand the given data
We have a thin convergent glass lens with:
- Refractive index of glass, \( \mu_g = 1.5 \)
- Power of the lens, \( P = +5.0 \, D \)

### Step 2: Calculate the focal length of the lens in air
The relationship between power \( P \) and focal length \( f \) is given by:
\[ P = \frac{1}{f} \quad \text{(in meters)} \]
Thus, we can find the focal length \( f \):
\[ f = \frac{1}{P} = \frac{1}{5.0} = 0.2 \, \text{m} = 20 \, \text{cm} \]

### Step 3: Use the lens maker's formula for the lens in air
The lens maker's formula is:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For our lens in air:
\[ \frac{1}{20} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
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 4: Analyze the lens when immersed 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 \( -100 \, \text{cm} \):
\[ \frac{1}{f} = \frac{\mu_g}{\mu_1} - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{-100} = \frac{1.5}{\mu_1} - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
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 5: Relate Equations (1) and (2)
From Equation (1):
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{10} \]
Substituting this into Equation (2):
\[ -\frac{1}{100} = \left( \frac{1.5}{\mu_1} - 1 \right) \left( \frac{1}{10} \right) \]
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 value of \( \mu_1 \) is:
\[ \mu_1 = \frac{5}{3} \]

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