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 use the lens maker's formula and the properties of mirrors.

### Step 1: Identify the given values
- Radii of curvature: \( R_1 = 20 \, \text{cm} \) (convex side), \( R_2 = -30 \, \text{cm} \) (concave side, negative due to sign convention)
- Refractive index of glass: \( \mu_g = 1.5 \)

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

### Step 3: Calculate \( \frac{1}{20} + \frac{1}{30} \)
Finding a common denominator (which is 60):
\[ \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60} \]
Thus,
\[ \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12} \]

### Step 4: Substitute back into the lens maker's formula
\[ \frac{1}{f_l} = 0.5 \times \frac{1}{12} = \frac{1}{24} \]
So,
\[ f_l = 24 \, \text{cm} \]

### Step 5: Determine the focal length of the silvered surface
Since the lens is silvered at the surface with the smaller radius of curvature (20 cm), it behaves like a concave mirror. The focal length of a mirror is given by:
\[ f_m = -\frac{R}{2} \]
For the silvered surface:
\[ f_m = -\frac{20}{2} = -10 \, \text{cm} \]

### Step 6: Combine the focal lengths of the lens and the mirror
The equivalent focal length \( f_{eq} \) of the combination of the lens and the 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 (which is 120):
\[ \frac{1}{24} = \frac{5}{120}, \quad \frac{1}{-10} = -\frac{12}{120} \]
Thus,
\[ \frac{1}{f_{eq}} = \frac{5 - 12}{120} = \frac{-7}{120} \]
So,
\[ f_{eq} = -\frac{120}{7} \approx -17.14 \, \text{cm} \]

### Step 7: Conclusion
Since the focal length is negative, the combination behaves like a concave lens (or mirror).

### Final Answer
The system behaves like a concave mirror with an equivalent focal length of approximately \( -17.14 \, \text{cm} \).

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