Question

Two convex lenses (\(L_1\) and \(L_2\)) of equal focal length \(f\) are placed at a distance \(\frac{f}{2}\) apart. An object is placed at a distance \(4f\) to the left of \(L_1\) as shown in the figure. The final image is at:

• A. \(\frac{5f}{11}\) right of \(L_2\)
• B. \(\frac{5f}{11}\) left of \(L_2\)
• C. \(5f\) right of \(L_2\)
• D. \(5f\) left of \(L_2\)

Hint:

For multiple lens systems, calculate the image formed by the first lens and then treat this image as the object for the next lens. Use the lens formula for each lens to find the final image position.

Answer 1

The Correct Option is A

Step 1: For the first lens (\(L_1\)), the object is placed at \(4f\) from \(L_1\). Using the lens formula: \[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} \] where \(u_1 = -4f\) (object distance for \(L_1\)) and \(v_1\) is the image distance from \(L_1\), we solve for \(v_1\): \[ \frac{1}{f} = \frac{1}{v_1} + \frac{1}{4f} \] \[ \frac{1}{v_1} = \frac{1}{f} - \frac{1}{4f} = \frac{3}{4f} \] So, \[ v_1 = \frac{4f}{3} \] The image formed by \(L_1\) is located \(\frac{4f}{3}\) to the right of \(L_1\).

Step 2: This image acts as the object for the second lens (\(L_2\)). The distance between the two lenses is \(\frac{f}{2}\), so the object for \(L_2\) is at a distance: \[ u_2 = \frac{4f}{3} - \frac{f}{2} = \frac{8f}{6} - \frac{3f}{6} = \frac{5f}{6} \] Now, using the lens formula for \(L_2\): \[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} \] where \(u_2 = \frac{5f}{6}\) and \(v_2\) is the image distance for \(L_2\), we solve for \(v_2\): \[ \frac{1}{f} = \frac{1}{v_2} - \frac{6}{5f} \] \[ \frac{1}{v_2} = \frac{1}{f} + \frac{6}{5f} = \frac{11}{5f} \] So, \[ v_2 = \frac{5f}{11} \] Therefore, the final image is at a distance of \(\frac{5f}{11}\) to the right of \(L_2\).