An object is placed 1m in front of the curved surface of a plano-convex lens whose plane surface is silvered. A real image is formed in front of the lens at a distance of 120cm. Then, the focal length of the lens is

Options

A. 100cm
B. 120cm
C. 109.1cm
D. 110.0cm

Explanation

To solve the problem, we need to find the focal length of a plano-convex lens with its plane surface silvered. The object is placed 1 meter (100 cm) in front of the lens, and a real image is formed at a distance of 120 cm in front of the lens.

### Step-by-Step Solution:

1. **Identify the given values:**
- Object distance (u) = -100 cm (negative because it is in front of the lens)
- Image distance (v) = -120 cm (negative because the image is formed in front of the lens)
- The lens is plano-convex with its plane surface silvered.

2. **Understand the behavior of the lens:**
- The plano-convex lens will refract the light at the curved surface and reflect it at the silvered plane surface.
- The image formed by the lens will act as a virtual object for the plane mirror (silvered surface).

3. **Apply the lens formula for the first surface (curved surface):**
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{u} \]
Here, \(v_1\) is the image distance formed by the lens before it reflects off the silvered surface. We will denote the focal length of the lens as \(f\).

4. **Rearranging the lens formula:**
\[ \frac{1}{v_1} = \frac{1}{f} + \frac{1}{100} \]
This gives us our first equation (Equation 1).

5. **Consider the second surface (silvered plane surface):**
- The image formed by the lens acts as a virtual object for the silvered surface. The distance of this virtual object from the silvered surface is \(v_1\).
- For the silvered surface, the object distance is equal to the image distance (as per the mirror formula):
\[ v_1 = -v_1 \]

6. **Apply the lens formula for the second surface:**
\[ \frac{1}{f} = \frac{1}{120} - \frac{1}{v_1} \]
Rearranging gives us:
\[ \frac{1}{v_1} = \frac{1}{120} - \frac{1}{f} \]
This gives us our second equation (Equation 2).

7. **Substitute Equation 1 into Equation 2:**
From Equation 1, we have:
\[ \frac{1}{v_1} = \frac{1}{f} + \frac{1}{100} \]
Substitute this into Equation 2:
\[ \frac{1}{f} + \frac{1}{100} = \frac{1}{120} - \frac{1}{f} \]

8. **Combine and solve for \(f\):**
\[ 2 \cdot \frac{1}{f} = \frac{1}{120} - \frac{1}{100} \]
Finding a common denominator (600):
\[ 2 \cdot \frac{1}{f} = \frac{5 - 6}{600} = -\frac{1}{600} \]
Thus,
\[ \frac{1}{f} = -\frac{1}{1200} \]
Therefore,
\[ f = -1200 \text{ cm} \]

9. **Final Calculation:**
Since the focal length is positive for a plano-convex lens, we take the absolute value:
\[ f = 1200 \text{ cm} \]

### Conclusion:
The focal length of the plano-convex lens is **120 cm**.