Two symmetric double convex lenses A and B have same focal length but the radii of curvature differ so that `R_(A) = 0.9R_(B)`. If refractive index of A is 1.63 find the refractive index of B.

Options

A. `1.7`
B. `1.6`
C. `1.5`
D. `4//3`

Explanation

To solve the problem, we will use the lens maker's formula, which relates the focal length of a lens to its refractive index and the radii of curvature of its surfaces.

### Step-by-Step Solution:

1. **Understanding the Lens Maker's Formula**:
The lens maker's formula is given by:
\[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a double convex lens, we can take \( R_1 = R \) (positive) and \( R_2 = -R \) (negative), leading to:
\[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R} + \frac{1}{R} \right) = \left( \mu - 1 \right) \left( \frac{2}{R} \right) \]
Thus, we can express the focal length as:
\[ f = \frac{R}{2(\mu - 1)} \]

2. **Setting Up the Equations for Lenses A and B**:
For lens A:
\[ f_A = \frac{R_A}{2(\mu_A - 1)} \]
For lens B:
\[ f_B = \frac{R_B}{2(\mu_B - 1)} \]
Given that \( R_A = 0.9 R_B \) and \( f_A = f_B \), we can equate the two focal lengths:
\[ \frac{0.9 R_B}{2(\mu_A - 1)} = \frac{R_B}{2(\mu_B - 1)} \]

3. **Canceling Common Terms**:
We can cancel \( R_B \) from both sides (assuming \( R_B \neq 0 \)):
\[ \frac{0.9}{2(\mu_A - 1)} = \frac{1}{2(\mu_B - 1)} \]

4. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 0.9(\mu_B - 1) = (\mu_A - 1) \]

5. **Substituting the Known Value**:
We know \( \mu_A = 1.63 \):
\[ 0.9(\mu_B - 1) = 1.63 - 1 \]
Simplifying the right side:
\[ 0.9(\mu_B - 1) = 0.63 \]

6. **Solving for \( \mu_B \)**:
Dividing both sides by 0.9:
\[ \mu_B - 1 = \frac{0.63}{0.9} \]
Calculating the right side:
\[ \mu_B - 1 = 0.7 \]
Therefore:
\[ \mu_B = 1.7 \]

### Final Answer:
The refractive index of lens B is \( \mu_B = 1.7 \).

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