A concave spherical surface of radius of curvature `10 cm` separates two mediums `X` and `Y` of refractive indices `4//3` and `3//2` respectively. Centre of curvature of the surfaces lies in the medium `X`. An object is placed in medium `X`.

Options

A. Image is always real
B. Image is real if the object distance is greater than `90 cm`
C. Image is always virtual
D. Image is virtual only If the object distance is less than `90 cm`.

Explanation

To solve the problem, we will use the lens maker's formula and the information given about the concave spherical surface separating two media.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of curvature (R) = -10 cm (negative because the center of curvature is in medium X).
- Refractive index of medium X (n1) = 4/3.
- Refractive index of medium Y (n2) = 3/2.

2. **Set Up the Lens Maker's Formula:**
The lens maker's formula for a spherical surface is given by:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
where:
- \( n_1 \) = refractive index of the medium where the object is placed (medium X).
- \( n_2 \) = refractive index of the medium where the image is formed (medium Y).
- \( u \) = object distance (we will take it as -x, where x is positive).
- \( v \) = image distance.

3. **Substitute the Values into the Formula:**
Substituting the values into the formula:
\[ \frac{3/2}{v} - \frac{4/3}{-x} = \frac{(3/2) - (4/3)}{-10} \]

4. **Simplify the Right Side:**
First, calculate \( \frac{3}{2} - \frac{4}{3} \):
\[ \frac{3}{2} = \frac{9}{6}, \quad \frac{4}{3} = \frac{8}{6} \quad \Rightarrow \quad \frac{3}{2} - \frac{4}{3} = \frac{9}{6} - \frac{8}{6} = \frac{1}{6} \]
Now substitute this into the equation:
\[ \frac{3/2}{v} + \frac{4/3}{x} = \frac{1/6}{-10} \]
Simplifying the right side gives:
\[ \frac{1/6}{-10} = -\frac{1}{60} \]

5. **Rearranging the Equation:**
Now we have:
\[ \frac{3/2}{v} + \frac{4/3}{x} = -\frac{1}{60} \]

6. **Finding a Common Denominator:**
Multiply through by \( 60vx \) to eliminate the denominators:
\[ 60x \cdot \frac{3}{2} + 60v \cdot \frac{4}{3} = -vx \]

7. **Solving for v:**
Rearranging gives:
\[ 90x + 80v = -vx \]
\[ 80v + vx = -90x \]
\[ v(80 + x) = -90x \]
\[ v = \frac{-90x}{80 + x} \]

8. **Analyzing the Sign of v:**
Since \( x \) is positive, \( v \) will be negative, indicating that the image is virtual.

### Conclusion:
The image formed is virtual.