Question

Two convex lenses A and B, each of focal length 10.0 cm, are mounted on an optical bench at 50.0 cm and 70.0 cm respectively. An object is mounted at 20.0 cm. Find the nature and position of the final image formed by the combination.

Hint:

To solve combined lens problems, first find the image formed by the first lens and use this as the object for the second lens.

Answer 1

Step 1: Image formation by the first lens. For lens A, the object distance is \( u_1 = -20.0 \, \text{cm} \) and the focal length is \( f_1 = 10.0 \, \text{cm} \). Applying the lens formula, \( \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1} \), we get \( \frac{1}{10.0} = \frac{1}{v_1} - \frac{1}{-20.0} \), which yields \( v_1 = 20.0 \, \text{cm} \). Therefore, lens A forms a real and inverted image at \( v_1 = 20.0 \, \text{cm} \).

Step 2: Image formation by the second lens. The image from lens A serves as the object for lens B. The object distance for lens B is \( u_2 = 70.0 - 20.0 = 50.0 \, \text{cm} \). Using the lens formula for lens B, \( \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2} \), we have \( \frac{1}{10.0} = \frac{1}{v_2} - \frac{1}{50.0} \), resulting in \( v_2 = 12.5 \, \text{cm} \). Consequently, the final image is formed 12.5 cm from lens B, and it is real and inverted.