Two converging glass lenses A and B have focal lengths in the ratio `2:1`. The radius of curvature of first surface of lens A is 1/4 th of the second surface where as the radius of curvature of first surface of lens B is twice that of second surface. Then the ratio between the radii of the first surfaces of A and B is

Options

A. `5:3`
B. `3:5`
C. `1:2`
D. `5:6`

Answer

To solve the problem, we need to find the ratio between the radii of the first surfaces of two converging lenses A and B, given their focal lengths and the relationships between their radii of curvature.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The focal lengths of lenses A and B are in the ratio \( f_A : f_B = 2 : 1 \).
- For lens A, the radius of curvature of the first surface \( r_{1A} \) is \( \frac{1}{4} \) of the second surface \( r_{2A} \).
- For lens B, the radius of curvature of the first surface \( r_{1B} \) is \( 2 \) times the second surface \( r_{2B} \).

2. **Use the Lens Maker's Formula**:
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{r_1} + \frac{1}{r_2} \right) \]
where \( \mu \) is the refractive index of the lens material.

3. **Apply the Formula for Lens A**:
- Given \( r_{1A} = \frac{1}{4} r_{2A} \), we can express \( r_{2A} \) as \( 4r_{1A} \).
- Substitute into the lens maker's formula:
\[ \frac{1}{f_A} = (\mu - 1) \left( \frac{1}{r_{1A}} + \frac{1}{4r_{1A}} \right) \]
\[ = (\mu - 1) \left( \frac{5}{4r_{1A}} \right) \]
Thus,
\[ f_A = \frac{4r_{1A}}{5(\mu - 1)} \]

4. **Apply the Formula for Lens B**:
- Given \( r_{1B} = 2r_{2B} \), we can express \( r_{2B} \) as \( \frac{1}{2} r_{1B} \).
- Substitute into the lens maker's formula:
\[ \frac{1}{f_B} = (\mu - 1) \left( \frac{1}{r_{1B}} + \frac{2}{r_{1B}} \right) \]
\[ = (\mu - 1) \left( \frac{3}{r_{1B}} \right) \]
Thus,
\[ f_B = \frac{r_{1B}}{3(\mu - 1)} \]

5. **Set Up the Ratio of Focal Lengths**:
We know \( \frac{f_A}{f_B} = \frac{2}{1} \):
\[ \frac{\frac{4r_{1A}}{5(\mu - 1)}}{\frac{r_{1B}}{3(\mu - 1)}} = 2 \]
The \( (\mu - 1) \) cancels out:
\[ \frac{4r_{1A}}{5} \cdot \frac{3}{r_{1B}} = 2 \]
Rearranging gives:
\[ 4r_{1A} \cdot 3 = 10r_{1B} \]
\[ 12r_{1A} = 10r_{1B} \]
\[ \frac{r_{1A}}{r_{1B}} = \frac{10}{12} = \frac{5}{6} \]

6. **Conclusion**:
The ratio of the radii of the first surfaces of lenses A and B is:
\[ \frac{r_{1A}}{r_{1B}} = \frac{5}{6} \]

### Final Answer:
The ratio between the radii of the first surfaces of A and B is \( \frac{5}{6} \).