The curvature radii of a concavo-convex glass lens are `20cm`and `60cm`. The convex surface of the lens is silvered .With the lens horizontal, the concave surface is filled with water. The focal length of the effective mirror is (`mu`of glass=`1.5,mu`of water=`4//3`

Options

A. `(90)/(13)cm`
B. `(80)/(13)cm`
C. `(20)/(3)cm`
D. `(45)/(8)cm`

Answer

To solve the problem, we need to find the effective focal length of a concavo-convex lens with given curvature radii and refractive indices. Here’s the step-by-step solution:

### Step 1: Identify the Parameters
We have a concavo-convex lens with the following parameters:
- Radius of curvature of the convex surface, \( R_1 = +60 \, \text{cm} \) (positive because it is convex)
- Radius of curvature of the concave surface, \( R_2 = -20 \, \text{cm} \) (negative because it is concave)
- Refractive index of glass, \( \mu_g = 1.5 \)
- Refractive index of water, \( \mu_w = \frac{4}{3} \)

### Step 2: Calculate the Focal Length of the Water Lens
Using the lens maker's formula for the water-filled concave surface:
\[ \frac{1}{F_w} = \left( \mu_w - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{F_w} = \left( \frac{4}{3} - 1 \right) \left( \frac{1}{\infty} - \frac{1}{-60} \right) \]
Since \( \frac{1}{\infty} = 0 \):
\[ \frac{1}{F_w} = \left( \frac{1}{3} \right) \left( 0 + \frac{1}{60} \right) = \frac{1}{3} \cdot \frac{1}{60} = \frac{1}{180} \]
Thus,
\[ F_w = 180 \, \text{cm} \]

### Step 3: Calculate the Focal Length of the Glass Lens
Using the lens maker's formula for the glass lens:
\[ \frac{1}{F_g} = \left( \mu_g - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{F_g} = \left( 1.5 - 1 \right) \left( \frac{1}{-60} - \frac{1}{-20} \right) \]
Calculating:
\[ \frac{1}{F_g} = \left( 0.5 \right) \left( -\frac{1}{60} + \frac{1}{20} \right) \]
Finding a common denominator:
\[ -\frac{1}{60} + \frac{3}{60} = \frac{2}{60} = \frac{1}{30} \]
Thus,
\[ \frac{1}{F_g} = 0.5 \cdot \frac{1}{30} = \frac{1}{60} \]
So,
\[ F_g = 60 \, \text{cm} \]

### Step 4: Calculate the Focal Length of the Mirror
The focal length of the silvered convex surface (mirror) is given by:
\[ F_m = \frac{R}{2} = \frac{20}{2} = 10 \, \text{cm} \]
Since it is a mirror, it will be negative:
\[ F_m = -10 \, \text{cm} \]

### Step 5: Calculate the Effective Focal Length
Using the formula for the effective focal length \( F \):
\[ \frac{1}{F} = 2 \cdot \frac{1}{F_w} + 2 \cdot \frac{1}{F_g} - \frac{1}{F_m} \]
Substituting the values:
\[ \frac{1}{F} = 2 \cdot \frac{1}{180} + 2 \cdot \frac{1}{60} - \left(-\frac{1}{10}\right) \]
Calculating each term:
\[ \frac{1}{F} = \frac{2}{180} + \frac{2}{60} + \frac{1}{10} \]
Finding a common denominator (180):
\[ \frac{1}{F} = \frac{2}{180} + \frac{6}{180} + \frac{18}{180} = \frac{26}{180} \]
Thus,
\[ F = \frac{180}{26} = \frac{90}{13} \, \text{cm} \]

### Final Answer
The effective focal length of the mirror is:
\[ F = \frac{90}{13} \, \text{cm} \]