Plane surface of a thin plano-convex lens is silvered. If a point object is placed on principal axis at a distance `60cm` from the lens and fial image is formed at a distance `30cm` from the lens on same side. What will be the distance of final imagge from the lens if the plane surface is not silvered.

Options

A. `120cm`
B. `180cm`
C. `30cm`
D. `90cm`

Explanation

To solve the problem step-by-step, let's break it down into manageable parts.

### Step 1: Understand the Setup
We have a plano-convex lens with its plane surface silvered. This effectively turns the lens into a mirror for light reflecting off the silvered surface. The object is placed 60 cm from the lens, and the final image is formed 30 cm from the lens on the same side as the object.

### Step 2: Use the Mirror Formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
where:
- \( f \) = focal length of the mirror (which we need to find)
- \( v \) = image distance (30 cm, but since it's on the same side as the object, we take it as -30 cm)
- \( u \) = object distance (60 cm, taken as -60 cm)

### Step 3: Substitute Values into the Mirror Formula
Substituting the values into the mirror formula:
\[ \frac{1}{f} = \frac{1}{-30} + \frac{1}{-60} \]

### Step 4: Calculate the Focal Length
Finding a common denominator (which is 60):
\[ \frac{1}{f} = -\frac{2}{60} - \frac{1}{60} = -\frac{3}{60} = -\frac{1}{20} \]
Thus, the focal length \( f \) is:
\[ f = -20 \text{ cm} \]

### Step 5: Relate Focal Length of the Lens
For a plano-convex lens, the focal length \( f \) is related to the focal length of the silvered lens as:
\[ f' = -\frac{f}{2} = -\frac{-20}{2} = 10 \text{ cm} \]
This means the effective focal length of the silvered lens is 10 cm.

### Step 6: Find the Focal Length of the Unsilvered Lens
The focal length of the unsilvered plano-convex lens is:
\[ f = 20 \text{ cm} \]

### Step 7: Use the Lens Formula for the Unsilvered Lens
Now we will use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( f = 20 \) cm, \( u = -60 \) cm:
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-60} \]
This simplifies to:
\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{60} \]

### Step 8: Solve for Image Distance \( v \)
Finding a common denominator (which is 60v):
\[ \frac{3v}{60v} = \frac{3}{60} + \frac{v}{60v} \]
This leads to:
\[ \frac{3v + 20}{60v} = 1 \]
Cross-multiplying gives:
\[ 3v + 20 = 60v \]
Rearranging gives:
\[ 60v - 3v = 20 \implies 57v = 20 \implies v = \frac{20}{57} \approx 0.35 \text{ cm} \]

### Final Answer
The distance of the final image from the lens if the plane surface is not silvered is approximately 0.35 cm.