A concavo-convex glass (index=1.5) lens has radii of curvatures 60cm and 40cm, respectively. Its convex surface is silvered , and its placed on a horizontal table with concave surface up. The concave surface is then filled with a liquid of index 2.0. The combination behaves like

Options

A. Concave mirror
B. Convex mirror
C. Flat mirror
D. Convex lens

Answer

To solve the problem step by step, we will analyze the behavior of the concavo-convex lens when its convex surface is silvered and filled with a liquid of a certain refractive index.

### Step 1: Identify the Lens and its Properties
We have a concavo-convex lens made of glass with a refractive index (n_glass) of 1.5. The radii of curvature are given as:
- R1 (convex surface) = +60 cm (positive for convex)
- R2 (concave surface) = -40 cm (negative for concave)

### Step 2: Determine the Focal Length of the Lens
The focal length (f) of a lens can be calculated using the lens maker's formula:
\[ \frac{1}{f} = \left( n - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{60} - \frac{1}{-40} \right) \]
Calculating the right side:
\[ \frac{1}{f} = 0.5 \left( \frac{1}{60} + \frac{1}{40} \right) \]
Finding a common denominator (120):
\[ \frac{1}{60} = \frac{2}{120}, \quad \frac{1}{40} = \frac{3}{120} \]
So,
\[ \frac{1}{f} = 0.5 \left( \frac{2 + 3}{120} \right) = 0.5 \left( \frac{5}{120} \right) = \frac{5}{240} = \frac{1}{48} \]
Thus, the focal length \( f \) is:
\[ f = 48 \text{ cm} \]

### Step 3: Analyze the Silvered Surface
The convex surface of the lens is silvered, which means it acts as a concave mirror. The focal length of a concave mirror is given by:
\[ f_{mirror} = -\frac{R}{2} \]
For the convex surface (acting as a concave mirror):
\[ f_{mirror} = -\frac{60}{2} = -30 \text{ cm} \]

### Step 4: Determine the Effective Focal Length
When the lens is placed on a horizontal table with the concave surface filled with a liquid of refractive index \( n_{liquid} = 2.0 \), we need to find the effective focal length of the combination (lens + mirror). The effective focal length \( f_{effective} \) can be calculated using:
\[ \frac{1}{f_{effective}} = \frac{1}{f_{lens}} + \frac{1}{f_{mirror}} \]
Substituting the values:
\[ \frac{1}{f_{effective}} = \frac{1}{48} + \frac{1}{-30} \]
Finding a common denominator (240):
\[ \frac{1}{48} = \frac{5}{240}, \quad \frac{1}{-30} = -\frac{8}{240} \]
Thus,
\[ \frac{1}{f_{effective}} = \frac{5 - 8}{240} = \frac{-3}{240} = -\frac{1}{80} \]
So, the effective focal length is:
\[ f_{effective} = -80 \text{ cm} \]

### Step 5: Conclusion
The negative effective focal length indicates that the combination behaves like a concave mirror.