Question

There are 3 lenses as shown in diagram of focal lengths (in cm) & distances depicted. The final image (real) is 30 cm from right-most lens then distance $d$ is

Object $f=10$ $f=10$ $f=30$

← 30 cm → ← d → ← 5 cm→

Answer 1

25 cm

Answer 2

1. Image formation by L1: Object distance $u_1 = -30$ cm, $f_1 = 10$ cm. Using $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$, $v_1 = 15$ cm.

2. Image formation by L2: Object distance for L2 is $u_2 = d - v_1 = d - 15$. Focal length $f_2 = -10$ cm. For parallel rays from L2 (object at focal point), $u_2 = -f_2 = 10$ cm. Thus, $d - 15 = 10 \implies d = 25$ cm.

3. Verification with L3: If $d=25$ cm, $u_2 = 10$ cm. For L2: $\frac{1}{v_2} - \frac{1}{10} = \frac{1}{-10} \implies v_2 = \infty$. Rays are parallel. For L3, $u_3 = \infty$, $f_3 = 30$ cm. $\frac{1}{v_3} - \frac{1}{\infty} = \frac{1}{30} \implies v_3 = 30$ cm. This matches the given condition.