An equiconcave lens having radius of curvature of each surface 20 cm has one surface silvered. If the refractive index of the lens is 1.5, then the magnitude of the focal length is

Options

A. 2.5cm
B. 0.4cm
C. 0
D. 5cm

Explanation

To solve the problem of finding the magnitude of the focal length of an equiconcave lens with one surface silvered, we will follow these steps:

### Step 1: Identify the parameters of the lens
- The radius of curvature (R) of each surface of the lens is given as 20 cm.
- The refractive index (μ) of the lens material is 1.5.

### Step 2: Calculate the focal length of the equiconcave lens
The formula for the focal length (f) of a lens is given by the lens maker's formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For an equiconcave lens:
- \( R_1 = -20 \) cm (the first surface is concave)
- \( R_2 = +20 \) cm (the second surface is also concave)

Substituting these values into the formula:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{-20} - \frac{1}{20} \right) \]
\[ \frac{1}{f} = 0.5 \left( -\frac{1}{20} - \frac{1}{20} \right) \]
\[ \frac{1}{f} = 0.5 \left( -\frac{2}{20} \right) = 0.5 \left( -\frac{1}{10} \right) = -\frac{1}{20} \]
Thus, the focal length \( f \) of the lens is:
\[ f = -20 \text{ cm} \]

### Step 3: Account for the silvered surface
When one surface of the lens is silvered, it acts as a combination of a lens and a mirror. The silvered surface acts as a convex mirror.

### Step 4: Calculate the focal length of the convex mirror
The focal length (f_m) of a convex mirror is given by:
\[ f_m = \frac{R}{2} \]
Where \( R \) is the radius of curvature of the mirror. Since the radius of curvature is 20 cm, we have:
\[ f_m = \frac{20}{2} = 10 \text{ cm} \]

### Step 5: Use the lens-mirror combination formula
For a lens and mirror combination, the equivalent focal length (f_eq) is given by:
\[ \frac{1}{f_{eq}} = \frac{1}{f_m} - \frac{2}{f_l} \]
Substituting the values:
\[ \frac{1}{f_{eq}} = \frac{1}{10} - \frac{2}{-20} \]
\[ \frac{1}{f_{eq}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \]
Thus, the equivalent focal length \( f_{eq} \) is:
\[ f_{eq} = 5 \text{ cm} \]

### Conclusion
The magnitude of the focal length of the lens with one surface silvered is:
\[ \boxed{5 \text{ cm}} \]