Question

A concave lens and a convex lens are arranged as shown in the figure. The position of the final image is:

• A. \(17\ \text{cm}\) to the left of convex lens
• B. \(24.2\ \text{cm}\) to the right of concave lens
• C. \(29.2\ \text{cm}\) to the right of concave lens
• D. \(24.2\ \text{cm}\) to the left of convex lens

Hint:

For a combination of lenses, first find the image formed by the first lens. This image acts as the object for the second lens. Always use proper sign convention carefully.

Answer 1

The Correct Option is C

Step 1: Image formed by the concave lens.
For the concave lens, \[ f_1=-20\ \text{cm} \] The object is \(30\ \text{cm}\) to the left of the concave lens, so \[ u_1=-30\ \text{cm} \] Using the lens formula, \[ \frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1} \] Substituting values, \[ \frac{1}{v_1}-\frac{1}{-30}=\frac{1}{-20} \] \[ \frac{1}{v_1}+\frac{1}{30}=-\frac{1}{20} \] \[ \frac{1}{v_1}=-\frac{1}{20}-\frac{1}{30} \] \[ \frac{1}{v_1}=-\frac{3+2}{60} \] \[ \frac{1}{v_1}=-\frac{5}{60} \] \[ v_1=-12\ \text{cm} \] So, the concave lens forms a virtual image \(12\ \text{cm}\) to its left.

Step 2: This image acts as object for the convex lens.
The distance between the concave lens and convex lens is \[ 5\ \text{cm} \] The image formed by the concave lens is \(12\ \text{cm}\) to the left of the concave lens.
Therefore, its distance from the convex lens is \[ 12+5=17\ \text{cm} \] So, for the convex lens, \[ u_2=-17\ \text{cm} \] Also, \[ f_2=10\ \text{cm} \]

Step 3: Apply lens formula for convex lens.
Using \[ \frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2} \] Substituting values, \[ \frac{1}{v_2}-\frac{1}{-17}=\frac{1}{10} \] \[ \frac{1}{v_2}+\frac{1}{17}=\frac{1}{10} \] \[ \frac{1}{v_2}=\frac{1}{10}-\frac{1}{17} \] \[ \frac{1}{v_2}=\frac{17-10}{170} \] \[ \frac{1}{v_2}=\frac{7}{170} \] \[ v_2=\frac{170}{7} \] \[ v_2=24.2\ \text{cm} \] Thus, the final image is \(24.2\ \text{cm}\) to the right of the convex lens.

Step 4: Find the position with respect to the concave lens.
Since the convex lens is \(5\ \text{cm}\) to the right of the concave lens, the final image is at a distance \[ 5+24.2=29.2\ \text{cm} \] to the right of the concave lens.

Step 5: Final conclusion.
Hence, the final image is \[ \boxed{29.2\ \text{cm}\ \text{to the right of concave lens}} \]