Two Lens System – Detailed Solution

Two Lens System Problem

Question (JEE Main 2023): A point object O is placed in front of two thin symmetrical coaxial convex lenses \(L_1\) and \(L_2\). The focal lengths of the lenses are \(f_1 = 12\,\text{cm}\) and \(f_2 = 11\,\text{cm}\), respectively. The distance between the two lenses is \(10\,\text{cm}\), and the object is placed \(6\,\text{cm}\) away from lens \(L_1\). As shown in the figure, find the distance (in cm) between the object and the final image formed by this two-lens system.

Select the Correct Option

Possible Answers:

(A) 28 cm

(B) 32 cm

(C) 38 cm

(D) 40 cm

Step-by-Step Explanation

1. Lens \(L_1\) (focal length \(12\,\text{cm}\)): Object is \(6\,\text{cm}\) from lens \(L_1\). Let \(u_1 = -6\,\text{cm}\) (using sign convention). Lens formula: \(\displaystyle \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}\). Substituting \(\displaystyle \frac{1}{v_1} - \frac{1}{-6} = \frac{1}{12}\). This gives \(\displaystyle \frac{1}{v_1} + \frac{1}{6} = \frac{1}{12}\). Hence, \(\displaystyle \frac{1}{v_1} = \frac{1}{12} - \frac{1}{6} = \frac{1}{12} - \frac{2}{12} = -\frac{1}{12}\). So \(v_1 = -12\,\text{cm}\).
   This means the image is formed on the same side as the object (virtual image) \(12\,\text{cm}\) away from lens \(L_1\).
2. Position of This Image Relative to Lens \(L_2\): The two lenses are separated by \(10\,\text{cm}\). So the distance from lens \(L_1\) to lens \(L_2\) is \(10\,\text{cm}\). The image from \(L_1\) is formed \emph{behind} \(L_1\), so effectively, lens \(L_2\) sees the object at a distance \(\displaystyle (10 + 12) = 22\,\text{cm}\) \emph{in front} of it. (Because the virtual image behind \(L_1\) acts like an object for \(L_2\).)
3. Lens \(L_2\) (focal length \(11\,\text{cm}\)): Let \(u_2 = -22\,\text{cm}\). Lens formula: \(\displaystyle \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}\). That is, \(\displaystyle \frac{1}{v_2} - \frac{1}{-22} = \frac{1}{11}\). Thus, \(\displaystyle \frac{1}{v_2} + \frac{1}{22} = \frac{1}{11} = \frac{2}{22}\). Therefore, \(\displaystyle \frac{1}{v_2} = \frac{2}{22} - \frac{1}{22} = \frac{1}{22}\). Hence, \(v_2 = 22\,\text{cm}\).
4. Distance Between the Object and Final Image: The final image is \(22\,\text{cm}\) from lens \(L_2\). Lens \(L_2\) is \(10\,\text{cm}\) from lens \(L_1\). The object is \(6\,\text{cm}\) in front of lens \(L_1\). So total distance = \(6 + 10 + 22 = 38\,\text{cm}\).
5. Final Answer: The distance between the object and the final image is \(38\,\text{cm}\).

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