A thin equiconvex 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 principle axis of the system such that its final image coincides with itself. Now few drops of a transparent liquid are placed between the mirror and lens such that final image of the object is at infinity. Find (a) the refractive index of the liquid used and (b the position of the object.

Answer

To solve the problem step by step, we will break it down into parts (a) and (b) as required.

### Step-by-Step Solution

**Given:**
- Refractive index of the lens, \( n_L = \frac{3}{2} \)
- Radius of curvature of the lens, \( R_L = 50 \, \text{cm} \)
- Radius of curvature of the mirror, \( R_M = 100 \, \text{cm} \)

#### Part (a): Finding the Refractive Index of the Liquid

1. **Calculate the Focal Length of the Lens:**
The formula for the focal length \( f_L \) of a lens is given by:
\[ \frac{1}{f_L} = (n_L - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For an equiconvex lens, \( R_1 = +50 \, \text{cm} \) and \( R_2 = -50 \, \text{cm} \):
\[ \frac{1}{f_L} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{50} - \left(-\frac{1}{50}\right) \right) \]
\[ = \frac{1}{2} \left( \frac{1}{50} + \frac{1}{50} \right) = \frac{1}{2} \cdot \frac{2}{50} = \frac{1}{50} \]
Thus, \( f_L = 50 \, \text{cm} \).

2. **Calculate the Power of the Lens:**
The power \( P_L \) of the lens is given by:
\[ P_L = \frac{1}{f_L} = \frac{1}{50} \, \text{cm}^{-1} \]

3. **Calculate the Power of the Mirror:**
The focal length of the mirror \( f_M \) is:
\[ f_M = -\frac{R_M}{2} = -\frac{100}{2} = -50 \, \text{cm} \]
Thus, the power \( P_M \) of the mirror is:
\[ P_M = \frac{1}{f_M} = -\frac{1}{50} \, \text{cm}^{-1} \]

4. **Calculate the Equivalent Power of the System:**
The equivalent power \( P_{EM} \) of the lens-mirror system is:
\[ P_{EM} = P_L + P_M = \frac{1}{50} - \frac{1}{50} = 0 \]

5. **Determine the Condition for the Final Image at Infinity:**
For the final image to be at infinity, the object must be at the focus of the equivalent mirror. Thus, the focal length of the equivalent mirror must be \( f_{EM} = 100 \, \text{cm} \).

6. **Set Up the Equation for the Liquid:**
The power of the liquid \( P_L \) can be expressed as:
\[ P_L = \frac{1}{f_L} = \frac{1}{-50} - \frac{1}{100} \]
We can express the power of the liquid in terms of its refractive index \( n \):
\[ P_L = 3 \left( \frac{n - 1}{100} \right) \]

7. **Equate the Powers:**
Set the power of the liquid equal to the required power:
\[ 0 = 2P_L + P_M \]
Substitute the values and solve for \( n \):
\[ 0 = 2 \left( 3 \left( \frac{n - 1}{100} \right) \right) - \frac{1}{50} \]
Simplifying gives:
\[ 6(n - 1) - 2 = 0 \implies 6n - 6 - 2 = 0 \implies 6n = 8 \implies n = \frac{4}{3} \]

**Final Result for Part (a):**
The refractive index of the liquid used is \( n = \frac{4}{3} \).

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#### Part (b): Finding the Position of the Object

1. **Determine the Position of the Object:**
Since the final image coincides with the object, the object must be at the center of curvature of the mirror:
\[ \text{Position of the object} = R_M = 100 \, \text{cm} \]

**Final Result for Part (b):**
The position of the object is \( 100 \, \text{cm} \) from the mirror.

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