Question:
A convergent lens is placed \(40\,\text{cm}\) to the right of a divergent lens of focal length \(15\,\text{cm}\). A parallel beam of light enters the divergent lens from the left, and the beam is again parallel when it emerges from the convergent lens. The focal length of the convergent lens is
A convergent lens is placed \(40\,\text{cm}\) to the right of a divergent lens of focal length \(15\,\text{cm}\). A parallel beam of light enters the divergent lens from the left, and the beam is again parallel when it emerges from the convergent lens. The focal length of the convergent lens is
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If rays emerge parallel from a converging lens, the object for that lens must be located at its focal point. Track the image formed by the first lens carefully and use it as the object for the second lens.
Updated On: Jun 18, 2026
- \(40\,\text{cm}\)
- \(25\,\text{cm}\)
- \(55\,\text{cm}\)
- \(27.5\,\text{cm}\)
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The Correct Option is C
Solution and Explanation
Step 1: Determine the image formed by the divergent lens.
The divergent lens has focal length \[ f_1=-15\,\text{cm}. \] A parallel beam incident on a diverging lens appears to diverge from its principal focus. Hence the image formed by the first lens is a virtual image at \[ 15\,\text{cm} \] to the left of the divergent lens.
Step 2: Find the object distance for the convergent lens.
The convergent lens is placed \[ 40\,\text{cm} \] to the right of the divergent lens. Therefore, the virtual image formed by the first lens is at a distance \[ 40+15=55\,\text{cm} \] to the left of the convergent lens. Thus, for the convergent lens, \[ u=-55\,\text{cm}. \]
Step 3: Use the condition for parallel emergent rays.
The rays emerging from the convergent lens are parallel. Therefore, the object for the convergent lens must lie at its first focal point. Hence, \[ u=-f_2. \] Therefore, \[ f_2=55\,\text{cm}. \]
Step 4: Final conclusion.
Thus, the focal length of the convergent lens is \[ \boxed{55\,\text{cm}} \]
The divergent lens has focal length \[ f_1=-15\,\text{cm}. \] A parallel beam incident on a diverging lens appears to diverge from its principal focus. Hence the image formed by the first lens is a virtual image at \[ 15\,\text{cm} \] to the left of the divergent lens.
Step 2: Find the object distance for the convergent lens.
The convergent lens is placed \[ 40\,\text{cm} \] to the right of the divergent lens. Therefore, the virtual image formed by the first lens is at a distance \[ 40+15=55\,\text{cm} \] to the left of the convergent lens. Thus, for the convergent lens, \[ u=-55\,\text{cm}. \]
Step 3: Use the condition for parallel emergent rays.
The rays emerging from the convergent lens are parallel. Therefore, the object for the convergent lens must lie at its first focal point. Hence, \[ u=-f_2. \] Therefore, \[ f_2=55\,\text{cm}. \]
Step 4: Final conclusion.
Thus, the focal length of the convergent lens is \[ \boxed{55\,\text{cm}} \]
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