In a medium of refractive index 1.6 and having a convex surface has a point object in it at a distance of 12 cm from the pole. The radius of curvature is 6 cm. Locate the image as seen from air

Options

A. A real image at 30 cm
B. A virtual image at 30 cm
C. A real image at 4.28 cm
D. A virtual image at 4.28 cm

Answer

To solve the problem of locating the image as seen from air when a point object is placed in a medium with a refractive index of 1.6, we will follow these steps:

### Step 1: Identify the Given Values
- Refractive index of the medium (μ1) = 1.6
- Refractive index of air (μ2) = 1.0
- Object distance (u) = -12 cm (negative because the object is on the same side as the incoming light)
- Radius of curvature (R) = +6 cm (positive for convex surface)

### Step 2: Use the Refraction Formula
The formula for refraction at a spherical surface when light travels from a denser medium to a rarer medium is given by:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_1 - \mu_2}{R} \]
Where:
- \( v \) is the image distance we need to find.
- \( u \) is the object distance (which we have as -12 cm).
- \( R \) is the radius of curvature (which we have as +6 cm).
- \( \mu_1 \) is the refractive index of the medium (1.6).
- \( \mu_2 \) is the refractive index of air (1.0).

### Step 3: Substitute the Values into the Formula
Substituting the values into the formula, we get:
\[ \frac{1}{v} - \frac{1.6}{-12} = \frac{1.6 - 1}{6} \]

### Step 4: Simplify the Equation
First, simplify the right side:
\[ \frac{1.6 - 1}{6} = \frac{0.6}{6} = 0.1 \]
Now, substituting this back into the equation:
\[ \frac{1}{v} + \frac{1.6}{12} = 0.1 \]
Calculating \( \frac{1.6}{12} \):
\[ \frac{1.6}{12} = 0.1333 \]
Now, the equation becomes:
\[ \frac{1}{v} + 0.1333 = 0.1 \]

### Step 5: Solve for \( \frac{1}{v} \)
Rearranging gives:
\[ \frac{1}{v} = 0.1 - 0.1333 = -0.0333 \]

### Step 6: Find the Image Distance \( v \)
Taking the reciprocal:
\[ v = \frac{1}{-0.0333} \approx -30 \text{ cm} \]

### Step 7: Interpret the Result
The negative sign indicates that the image is formed on the same side as the object, which means it is a virtual image.

### Final Answer
The image is located 30 cm from the convex surface on the same side as the object, indicating it is a virtual image.

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