The flat face of a plano-convex lens of focal length 10 cm is silvered. A point source placed 30 cm in front of the curved surface will produce a

Options

A. real image 15 cm away from the lens
B. real image 6 cm away from the lens
C. virtual image 15 cm away from the lens
D. virtual image 6 cm away from the lens

Explanation

To solve the problem step by step, we will analyze the situation involving a plano-convex lens with its flat face silvered, and a point source placed in front of the curved surface.

### Step 1: Understand the System
We have a plano-convex lens with a focal length (f) of 10 cm. The flat face of the lens is silvered, which effectively turns the lens into a combination of a lens and a mirror. The point source is placed 30 cm in front of the curved surface of the lens.

### Step 2: Identify the Focal Length of the System
When the flat face of the lens is silvered, we need to consider the effective focal length of the system. The focal length of the plano-convex lens is given as:
- Focal length of the lens (f_lens) = +10 cm (positive because it is a converging lens)

When the flat face is silvered, it acts like a concave mirror. The focal length of a concave mirror is negative, and it can be calculated as:
- Focal length of the mirror (f_mirror) = -f_lens = -10 cm

### Step 3: Use the Lens-Mirror Formula
We can apply the lens formula for the system:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
Where:
- \( v \) = image distance from the lens
- \( u \) = object distance from the lens (negative since the object is in front of the lens)
- \( f \) = effective focal length of the system

Since the object is placed 30 cm in front of the curved surface, the object distance (u) is:
\[ u = -30 \text{ cm} \]

### Step 4: Calculate the Effective Focal Length
For the system, the effective focal length (f) can be derived from the lens and mirror combination. The effective focal length of the silvered plano-convex lens can be calculated as:
\[ \frac{1}{f} = \frac{1}{f_{lens}} + \frac{1}{f_{mirror}} \]
Substituting the values:
\[ \frac{1}{f} = \frac{1}{10} + \frac{1}{-10} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{10} - \frac{1}{10} = 0 \]
This indicates that the effective focal length of the system is infinite, which means it behaves like a plane mirror.

### Step 5: Substitute Values into the Lens Formula
Now, substituting the values into the lens formula:
\[ \frac{1}{v} + \frac{1}{-30} = 0 \]
This simplifies to:
\[ \frac{1}{v} = \frac{1}{30} \]
Thus,
\[ v = 30 \text{ cm} \]

### Step 6: Determine the Nature of the Image
The positive value of \( v \) indicates that the image is formed on the same side as the object, which means it is a real image. The image distance is 30 cm from the lens, which is the same distance as the object.

### Conclusion
The image formed is a real image located 30 cm from the lens.