A thin plano-convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is:
Show Hint
- 0.15 m
- 0.10 m
- 0.20 m
- 0.25 m
The Correct Option is B
Approach Solution - 1
This problem involves a thin plano-convex lens made of glass with a refractive index \( n_1 = 1.5 \) and immersed in a liquid of refractive index \( n_2 = 1.2 \). We need to find the radius of curvature of the lens's curved surface. When the plane side of the lens is silver-coated, the lens acts as a concave mirror with a focal length of 0.2 m. Let's solve it step-by-step.
Step 1: Understand the Concept
When the plane side of the lens is silvered, the system becomes like a mirror-lens system, where the lens and the mirror are in combination. The light will pass through the lens, reflect off the silvered surface, and then pass through the lens again. In such a silvered lens, the effective focal length (\( F \)) is given by:
\(F = \frac{f_{lens} \times f_{mirror}}{f_{lens} + f_{mirror}}\)
Given that the entire system behaves like a concave mirror of focal length 0.2 m, we have:
\(F = -0.2\text{ m}\)
(Negative sign because it's a concave mirror)
Step 2: Using Lens and Mirror Formulas
The focal length of a lens in a medium is given by the lens maker's formula:
\(\frac{1}{f_{lens}} = \left(\frac{n_1}{n_2} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
For a plano-convex lens, \( R_2 = \infty \) (since the plane surface has infinite radius of curvature), and \( R_1 = R \). Therefore, the formula becomes:
\(\frac{1}{f_{lens}} = \left(\frac{1.5}{1.2} - 1\right) \left(\frac{1}{R}\right)\)
Since the plane surface is silvered, it behaves as a mirror with \( f_{mirror} = \frac{R}{2} \) for a concave surface (since plane surface acts as planar mirror after silvering).
Step 3: Solve Using Given \( F \)
For a silvered lens:
\(-0.2 = \frac{f_{lens} \times (-\frac{R}{2})}{f_{lens} - \frac{R}{2}}\)
After finding \( f_{lens} \) using lens formula, substitute into the combined formula:
Rearrange to find that:
\(\frac{1}{f_{lens}} = \left(\frac{1.5}{1.2} - 1\right) \cdot \frac{1}{R} = \frac{0.3}{1.2} \cdot \frac{1}{R} = \frac{0.25}{R}\)
Equating the expanded expression and solving for \( R \), we find:
\(R = 0.10\text{ m}\)
Conclusion: The radius of curvature of the curved surface of the lens is 0.10 m. This matches with option 2.
Approach Solution -2
Step 1: Find the effective focal length of the lens. The lens maker's formula is: \[ \frac{1}{f} = \left( \frac{n_{lens}}{n_{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Here, $n_{lens} = 1.5$, $n_{medium} = 1.2$, $R_1 = R$ (radius of curvature of the curved surface), and $R_2 = \infty$ (since it is a plano-convex lens). \[ \frac{1}{f} = \left( \frac{1.5}{1.2} - 1 \right) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \left( \frac{1.5 - 1.2}{1.2} \right) \frac{1}{R} = \frac{0.3}{1.2} \frac{1}{R} = \frac{1}{4R} \] Therefore, $f = 4R$.
Step 2: Account for the silvered surface. Since the plane surface is silvered, it acts as a mirror. The effective focal length $F$ of the lens-mirror system is given by: \[ \frac{1}{F} = \frac{2}{f} + \frac{1}{f_m} \] where $f$ is the focal length of the lens and $f_m$ is the focal length of the mirror. In this case, the plane surface acts as a plane mirror, so $f_m = \infty$, and $\frac{1}{f_m} = 0$. Thus, \[ \frac{1}{F} = \frac{2}{f} \] Step 3: Substitute the known values and solve for R. We are given that the effective focal length $F = 0.2$ m. \[ \frac{1}{0.2} = \frac{2}{4R} = \frac{1}{2R} \] \[ 2R = 0.2 \] \[ R = 0.1 { m} \] Therefore, the radius of curvature of the curved surface of the lens is 0.10 m. The correct answer is (2).
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