A convex lens of radii of curvature 20cm and 30 cm respectively. It is silvered at the surface which has smaller radius of curvature. Then it will behave as `(mu_(g) = 1.5)`

Options

A. concave mirror with equivalent focal length `30/11 cm`.
B. concave mirror with equivalent focal length `60/11 cm`.
C. convex mirror with equivalent focal length `30/11 cm`.
D. convex mirror with equivalent focal length `60/11 cm`.

Explanation

To solve the problem step by step, we will follow the principles of optics related to lenses and mirrors. Here’s the detailed solution:

### Step 1: Identify the given data
- The radii of curvature of the convex lens are:
- \( R_1 = 20 \, \text{cm} \) (for the first surface)
- \( R_2 = -30 \, \text{cm} \) (for the second surface, negative because it is a convex lens)
- The refractive index of the glass is \( \mu_g = 1.5 \).

### Step 2: Use the lens maker's formula
The lens maker's formula for a lens is given by:
\[ \frac{1}{F_L} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{F_L} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-30} \right) \]
This simplifies to:
\[ \frac{1}{F_L} = 0.5 \left( \frac{1}{20} + \frac{1}{30} \right) \]

### Step 3: Calculate the terms inside the parentheses
First, find a common denominator for \( \frac{1}{20} \) and \( \frac{1}{30} \):
- The least common multiple of 20 and 30 is 60.
- Thus, we have:
\[ \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60} \]
Adding these gives:
\[ \frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} \]

### Step 4: Substitute back into the lens formula
Now substitute back into the lens maker's formula:
\[ \frac{1}{F_L} = 0.5 \cdot \frac{5}{60} = \frac{5}{120} = \frac{1}{24} \]
Thus, the focal length of the lens is:
\[ F_L = 24 \, \text{cm} \]

### Step 5: Find the focal length of the silvered surface
Since the lens is silvered at the surface with the smaller radius of curvature (20 cm), we treat it as a plane mirror. The focal length of a plane mirror is given by:
\[ F_M = -\frac{R}{2} = -\frac{20}{2} = -10 \, \text{cm} \]

### Step 6: Calculate the equivalent focal length
The equivalent focal length \( F_{eq} \) of the system (lens + mirror) is given by:
\[ \frac{1}{F_{eq}} = \frac{1}{F_L} + \frac{1}{F_M} \]
Substituting the values:
\[ \frac{1}{F_{eq}} = \frac{1}{24} + \frac{1}{-10} \]
Finding a common denominator (120):
\[ \frac{1}{F_{eq}} = \frac{5}{120} - \frac{12}{120} = \frac{-7}{120} \]
Thus, the equivalent focal length is:
\[ F_{eq} = -\frac{120}{7} \approx -17.14 \, \text{cm} \]

### Step 7: Interpret the result
Since the equivalent focal length is negative, the system behaves like a concave mirror.

### Final Answer
The system behaves like a concave mirror with an equivalent focal length of \( \frac{-120}{7} \, \text{cm} \).