Question:
A concave lens of focal length \(10\) cm is cut into two identical plano-concave lenses. The focal length of each lens will be
A concave lens of focal length \(10\) cm is cut into two identical plano-concave lenses. The focal length of each lens will be
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When a symmetrical lens is cut perpendicular to its principal axis, the power of each part becomes half.
\[
P'=\frac{P}{2}
\qquad\Rightarrow\qquad
f'=2f
\]
in magnitude.
Updated On: Jul 2, 2026
- \(20\) cm
- \(30\) cm
- \(40\) cm
- \(5\) cm
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The Correct Option is A
Solution and Explanation
Concept:
The focal length of a lens is determined by the lens maker's formula,
\[
\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
\]
where
• \(f\) is the focal length,
• \(\mu\) is the refractive index,
• \(R_1\) and \(R_2\) are the radii of curvature. When a symmetric biconcave lens is cut into two equal parts by a plane perpendicular to the principal axis, each part becomes a plano-concave lens. The power of each half becomes half of the original power.
Step 1: Calculate the original power of the lens.
The focal length of the original lens is \[ f=-10\text{ cm} \] Hence, its power is \[ P=\frac{1}{f} = \frac{1}{-10} = -\frac{1}{10}\text{ cm}^{-1}. \]
Step 2: Determine the power of each half lens.
Since each half has only one curved surface, its power becomes half of the original power. Therefore, \[ P'=\frac{P}{2} \] \[ P'=\frac{-1/10}{2} = -\frac{1}{20}\text{ cm}^{-1}. \] Hence, \[ f'=-20\text{ cm}. \] Therefore, the magnitude of the focal length is \[ \boxed{20\text{ cm}} \]
• \(f\) is the focal length,
• \(\mu\) is the refractive index,
• \(R_1\) and \(R_2\) are the radii of curvature. When a symmetric biconcave lens is cut into two equal parts by a plane perpendicular to the principal axis, each part becomes a plano-concave lens. The power of each half becomes half of the original power.
Step 1: Calculate the original power of the lens.
The focal length of the original lens is \[ f=-10\text{ cm} \] Hence, its power is \[ P=\frac{1}{f} = \frac{1}{-10} = -\frac{1}{10}\text{ cm}^{-1}. \]
Step 2: Determine the power of each half lens.
Since each half has only one curved surface, its power becomes half of the original power. Therefore, \[ P'=\frac{P}{2} \] \[ P'=\frac{-1/10}{2} = -\frac{1}{20}\text{ cm}^{-1}. \] Hence, \[ f'=-20\text{ cm}. \] Therefore, the magnitude of the focal length is \[ \boxed{20\text{ cm}} \]
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