To solve the problem step by step, we will follow the reasoning laid out in the video transcript.
### Step 1: Understand the System
The problem states that a thin plano-convex lens acts like a concave mirror when silvered on its plane surface. The focal length of the system is given as \( f = 0.2 \, \text{m} \) or \( 20 \, \text{cm} \). The refractive index of the lens material is given as \( \mu = 1.5 \).
### Step 2: Relate the Focal Lengths
The focal length of the system can be expressed as:
\[
\frac{1}{f} = \frac{1}{f_{\text{lens}}} + \frac{1}{f_{\text{mirror}}}
\]
Since the plane mirror has an infinite focal length, we can simplify this to:
\[
\frac{1}{f} = \frac{2}{f_{\text{lens}}}
\]
Thus, we can rearrange this to find the focal length of the lens:
\[
f_{\text{lens}} = 2f
\]
Substituting \( f = 20 \, \text{cm} \):
\[
f_{\text{lens}} = 2 \times 20 \, \text{cm} = 40 \, \text{cm}
\]
### Step 3: Use the Lensmaker's Formula
The lensmaker's formula is given by:
\[
\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
\]
For a plano-convex lens, we have:
- \( R_1 = R \) (the radius of curvature of the convex surface)
- \( R_2 = \infty \) (the plane surface)
Substituting these into the lensmaker's formula gives:
\[
\frac{1}{f_{\text{lens}}} = (\mu - 1) \left( \frac{1}{R} - 0 \right)
\]
This simplifies to:
\[
\frac{1}{f_{\text{lens}}} = (\mu - 1) \frac{1}{R}
\]
### Step 4: Substitute Known Values
Now substituting the known values:
\[
\frac{1}{40 \, \text{cm}} = (1.5 - 1) \frac{1}{R}
\]
This simplifies to:
\[
\frac{1}{40 \, \text{cm}} = 0.5 \frac{1}{R}
\]
### Step 5: Solve for R
Rearranging gives:
\[
\frac{1}{R} = \frac{1}{40 \, \text{cm}} \times 2
\]
\[
\frac{1}{R} = \frac{2}{40 \, \text{cm}} = \frac{1}{20 \, \text{cm}}
\]
Thus,
\[
R = 20 \, \text{cm}
\]
### Step 6: Convert to Meters
Since the question asks for the radius of curvature in meters:
\[
R = 20 \, \text{cm} = 0.2 \, \text{m}
\]
### Final Answer
The radius of curvature of the convex surface of the lens is \( R = 0.2 \, \text{m} \).
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