A denser medium of refractive index 1.5 has a concave surface of radius of curvature 12 cm. An object is situated in the denser medium at a distance of 9 cm from the pole. Locate the image due to refraction in air.

Options

A. A real image at 8 cm
B. A virtual image at 8 cm
C. A real image at 4.8 cm
D. A virtual image at 4.8 cm

Answer

To solve the problem, we need to locate the image formed by a concave surface in a denser medium (refractive index 1.5) when an object is placed at a distance of 9 cm from the pole of the surface. The radius of curvature of the concave surface is given as 12 cm. We will use the formula for refraction at a spherical surface.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Refractive index of the denser medium (μ1) = 1.5
- Refractive index of air (μ2) = 1.0
- Radius of curvature (R) = -12 cm (negative because it is a concave surface)
- Object distance (U) = -9 cm (negative as per the sign convention for real objects)

2. **Use the Refraction Formula:**
The formula for refraction at a spherical surface is given by:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
Substituting the known values:
\[ \frac{1}{V} - \frac{1.5}{-9} = \frac{1 - 1.5}{-12} \]

3. **Simplify the Equation:**
The equation becomes:
\[ \frac{1}{V} + \frac{1.5}{9} = \frac{-0.5}{-12} \]
This simplifies to:
\[ \frac{1}{V} + \frac{1.5}{9} = \frac{0.5}{12} \]

4. **Calculate Each Term:**
- Calculate \(\frac{1.5}{9}\):
\[ \frac{1.5}{9} = \frac{1.5 \times 12}{9 \times 12} = \frac{18}{108} = \frac{1}{6} \]
- Calculate \(\frac{0.5}{12}\):
\[ \frac{0.5}{12} = \frac{0.5 \times 6}{12 \times 6} = \frac{3}{72} = \frac{1}{24} \]

5. **Combine the Terms:**
Now, substituting back:
\[ \frac{1}{V} + \frac{1}{6} = \frac{1}{24} \]
Rearranging gives:
\[ \frac{1}{V} = \frac{1}{24} - \frac{1}{6} \]

6. **Finding a Common Denominator:**
The common denominator for 24 and 6 is 24:
\[ \frac{1}{6} = \frac{4}{24} \]
Thus:
\[ \frac{1}{V} = \frac{1}{24} - \frac{4}{24} = -\frac{3}{24} = -\frac{1}{8} \]

7. **Calculate V:**
Therefore:
\[ V = -8 \text{ cm} \]

8. **Interpret the Result:**
The negative sign indicates that the image is virtual and located on the same side as the object.

### Final Answer:
The image is located at a distance of 8 cm from the pole on the same side as the object, and it is a virtual image.