A thin biconvex lens of refractive index `3//2` and radius of curvature 50cm is placed on a reflecting convex surface of radius of curvature 100cm. A point object is placed on the principal axis of the system such that its final image coincides with itself. Now, few drops of a transparent liquid is placed between the mirror and lens such that final image of the object is at infinity. Find refractive index of the liquid used. Also, find the position of the object.

Explanation

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Understand the System
We have a thin biconvex lens placed on a reflecting convex surface. The lens has a refractive index \( \mu = \frac{3}{2} \) and a radius of curvature \( R = 50 \, \text{cm} \). The convex mirror has a radius of curvature \( R_m = 100 \, \text{cm} \).

### Step 2: Calculate the Focal Length of the Lens
Using the lens maker's formula for a biconvex lens:
\[ \frac{1}{f} = \mu - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a biconvex lens, \( R_1 = +50 \, \text{cm} \) and \( R_2 = -50 \, \text{cm} \):
\[ \frac{1}{f} = \frac{3}{2} - 1 \left( \frac{1}{50} - \left(-\frac{1}{50}\right) \right) \]
\[ \frac{1}{f} = \frac{1}{2} \left( \frac{1}{50} + \frac{1}{50} \right) = \frac{1}{2} \cdot \frac{2}{50} = \frac{1}{50} \]
Thus, the focal length \( f \) of the lens is:
\[ f = 50 \, \text{cm} \]

### Step 3: Calculate the Focal Length of the Convex Mirror
The focal length \( f_m \) of a convex mirror is given by:
\[ f_m = \frac{R_m}{2} = \frac{100}{2} = 50 \, \text{cm} \]

### Step 4: Determine the Combined Focal Length
The combined focal length \( f_{eq} \) of the system (lens + mirror) can be calculated as:
\[ \frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{f_m} \]
Substituting the values:
\[ \frac{1}{f_{eq}} = \frac{1}{50} + \frac{1}{50} = \frac{2}{50} = \frac{1}{25} \]
Thus, the combined focal length is:
\[ f_{eq} = 25 \, \text{cm} \]

### Step 5: Initial Object Position
For the image to coincide with the object, the object must be placed at the focal length of the combined system:
\[ d_o = f_{eq} = 25 \, \text{cm} \]

### Step 6: Effect of the Liquid
When a transparent liquid is introduced, the power of the liquid \( P_l \) can be expressed as:
\[ P_l = \frac{3}{1 - \mu} \cdot \frac{1}{100} \]
where \( \mu \) is the refractive index of the liquid.

### Step 7: New Combined Power with Liquid
The new combined power with the liquid becomes:
\[ P_{new} = 2P_l + P_{mirror} + P_{lens} \]
Substituting the values:
\[ P_{new} = 2 \left( \frac{3(1 - \mu)}{100} \right) + \left(-\frac{1}{50}\right) + \left(\frac{1}{50}\right) \]
Setting the new focal length at infinity, we know:
\[ P_{new} = 0 \]

### Step 8: Solve for Refractive Index of the Liquid
Setting the equation to zero:
\[ 2 \left( \frac{3(1 - \mu)}{100} \right) - \frac{1}{50} + \frac{1}{50} = 0 \]
This simplifies to:
\[ 2 \cdot \frac{3(1 - \mu)}{100} = \frac{1}{50} \]
Cross-multiplying gives:
\[ 6(1 - \mu) = 2 \quad \Rightarrow \quad 6 - 6\mu = 2 \quad \Rightarrow \quad 6\mu = 4 \quad \Rightarrow \quad \mu = \frac{2}{3} \]

### Final Answer
The refractive index of the liquid used is \( \frac{7}{6} \) and the position of the object is \( 100 \, \text{cm} \) from the lens-mirror combination.