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: The object is \(4f\) away from the first lens (\(L_1\)). Applying the lens formula: \[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} \] where \(u_1 = -4f\) and \(v_1\) is the image distance from \(L_1\), we find \(v_1\): \[ \frac{1}{f} = \frac{1}{v_1} + \frac{1}{4f} \] \[ \frac{1}{v_1} = \frac{1}{f} - \frac{1}{4f} = \frac{3}{4f} \] Thus, \[ v_1 = \frac{4f}{3} \] The image from \(L_1\) is \(\frac{4f}{3}\) to the right of \(L_1\).

Step 2: This image is the object for the second lens (\(L_2\)). The lenses are \(\frac{f}{2}\) apart, so the object distance for \(L_2\) is: \[ u_2 = \frac{4f}{3} - \frac{f}{2} = \frac{8f}{6} - \frac{3f}{6} = \frac{5f}{6} \] Applying 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 \(v_2\): \[ \frac{1}{f} = \frac{1}{v_2} - \frac{6}{5f} \] \[ \frac{1}{v_2} = \frac{1}{f} + \frac{6}{5f} = \frac{11}{5f} \] Therefore, \[ v_2 = \frac{5f}{11} \] The final image is \(\frac{5f}{11}\) to the right of \(L_2\).