Question:
An object O is placed in front of two thin coaxial convex lenses A and B of focal lengths 24 cm and 9 cm respectively. The object is 6 cm to the left of lens A. If the final image is formed 18 cm to the right of lens B, find the separation between the lenses. (Lens A is placed left of lens B)
An object O is placed in front of two thin coaxial convex lenses A and B of focal lengths 24 cm and 9 cm respectively. The object is 6 cm to the left of lens A. If the final image is formed 18 cm to the right of lens B, find the separation between the lenses. (Lens A is placed left of lens B)
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For multi-lens systems, treat the image of first lens as object for second lens carefully with sign convention.
Updated On: Jun 19, 2026
- 5 cm
- 10 cm
- 8 cm
- 12 cm
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The Correct Option is B
Solution and Explanation
Step 1: Image by lens A.
Using lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] For lens A: \[ f_A = 24\,cm,\quad u_A = -6\,cm \] \[ \frac{1}{24} = \frac{1}{v_A} + \frac{1}{6} \]
Step 2: Solve for \(v_A\).
\[ \frac{1}{v_A} = \frac{1}{24} - \frac{1}{6} = \frac{1 - 4}{24} = -\frac{3}{24} \] \[ v_A = -8\,cm \]
Step 3: Image acts as object for lens B.
Let separation be \(d\). Then object distance for lens B: \[ u_B = -(d - 8) \]
Step 4: Apply lens B formula.
Final image is at \(v_B = +18\,cm\), \(f_B = 9\,cm\): \[ \frac{1}{9} = \frac{1}{18} - \frac{1}{u_B} \]
Step 5: Solve for separation.
\[ \frac{1}{u_B} = \frac{1}{18} - \frac{1}{9} = -\frac{1}{18} \Rightarrow u_B = -18 \] So: \[ -(d - 8) = -18 \Rightarrow d = 10\,cm \]
Step 6: Final conclusion.
Thus, separation between lenses is \(10\,cm\).
Final Answer: \[ \boxed{10\,cm} \]
Using lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] For lens A: \[ f_A = 24\,cm,\quad u_A = -6\,cm \] \[ \frac{1}{24} = \frac{1}{v_A} + \frac{1}{6} \]
Step 2: Solve for \(v_A\).
\[ \frac{1}{v_A} = \frac{1}{24} - \frac{1}{6} = \frac{1 - 4}{24} = -\frac{3}{24} \] \[ v_A = -8\,cm \]
Step 3: Image acts as object for lens B.
Let separation be \(d\). Then object distance for lens B: \[ u_B = -(d - 8) \]
Step 4: Apply lens B formula.
Final image is at \(v_B = +18\,cm\), \(f_B = 9\,cm\): \[ \frac{1}{9} = \frac{1}{18} - \frac{1}{u_B} \]
Step 5: Solve for separation.
\[ \frac{1}{u_B} = \frac{1}{18} - \frac{1}{9} = -\frac{1}{18} \Rightarrow u_B = -18 \] So: \[ -(d - 8) = -18 \Rightarrow d = 10\,cm \]
Step 6: Final conclusion.
Thus, separation between lenses is \(10\,cm\).
Final Answer: \[ \boxed{10\,cm} \]
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