The radii of curvatureof double convex lens of glass `(mu = 1.5)` are in the ratio of `1 : 2`. This lens renders the rays parallel coming from an illuminated filament at a distance of `6 cm`. Calculate the radii of curvature of its surfaces.

Explanation

To solve the problem of finding the radii of curvature of a double convex lens with a refractive index of \( \mu = 1.5 \) and a focal length of \( f = 6 \, \text{cm} \), we can follow these steps:

### Step 1: Understand the relationship between the radii of curvature
Given that the radii of curvature \( R_1 \) and \( R_2 \) are in the ratio of \( 1:2 \), we can express them as:
\[ R_1 = R \quad \text{and} \quad R_2 = 2R \]

### Step 2: Use the lens maker's formula
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values we know:
- \( f = 6 \, \text{cm} \)
- \( \mu = 1.5 \)

We can rewrite the formula:
\[ \frac{1}{6} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{1}{6} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 3: Substitute \( R_2 \) in terms of \( R_1 \)
Using \( R_2 = 2R_1 \):
\[ \frac{1}{6} = 0.5 \left( \frac{1}{R_1} - \frac{1}{2R_1} \right) \]
This can be simplified:
\[ \frac{1}{6} = 0.5 \left( \frac{2 - 1}{2R_1} \right) = 0.5 \left( \frac{1}{2R_1} \right) = \frac{1}{4R_1} \]

### Step 4: Solve for \( R_1 \)
Now, we can rearrange the equation to find \( R_1 \):
\[ \frac{1}{6} = \frac{1}{4R_1} \implies 4R_1 = 6 \implies R_1 = \frac{6}{4} = 1.5 \, \text{cm} \]

### Step 5: Calculate \( R_2 \)
Now, using \( R_1 \) to find \( R_2 \):
\[ R_2 = 2R_1 = 2 \times 1.5 = 3 \, \text{cm} \]

### Step 6: Final results
Thus, the radii of curvature are:
\[ R_1 = 1.5 \, \text{cm}, \quad R_2 = 3 \, \text{cm} \]

### Summary
The radii of curvature of the double convex lens are:
- \( R_1 = 1.5 \, \text{cm} \)
- \( R_2 = 3 \, \text{cm} \)