A convex lens A of focal length $20 ext{ cm}$ and a concave lens B of focal length $5 ext{ cm}$ are kept alo

Question

A convex lens A of focal length $20	ext{ cm}$ and a concave lens B of focal length $5	ext{ cm}$ are kept along the same axis with the distance $d$ between them. If a parallel beam of light falling on A leaves B as a parallel beam, then distance $d$ in cm will be:

Accepted Answer

1.  **Analysis of Lens A (Convex):** A parallel beam of light incident on a convex lens converges at its principal focus. Therefore, lens A attempts to form an image at a distance $v_A = f_A = +20	ext{ cm}$ from its optical center.
2.  **Analysis of Lens B (Concave):** For the light rays emerging from lens B to be parallel, the object for lens B must be located at its focal point (or effectively, the rays entering B must be directed towards its focus on the other side).
3.  **Afocal System Condition:** When a combination of lenses receives parallel light and emits parallel light, the separation distance $d$ is the algebraic sum of their focal lengths: $d = f_A + f_B$.
4.  **Calculation:**
    *   $f_A = +20	ext{ cm}$ (Convex)
    *   $f_B = -5	ext{ cm}$ (Concave)
    *   $d = 20 + (-5) = 15	ext{ cm}$.
5.  **Alternative Method (Lens Formula):** The image formed by A acts as a virtual object for B. Let the distance be $d$.
    *   Object distance for B: $u_B = +(20 - d)$ (since the rays are converging towards a point $20	ext{ cm}$ from A).
    *   Final image distance for B: $v_B = \infty$ (parallel beam).
    *   Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for lens B:
        $\frac{1}{\infty} - \frac{1}{20 - d} = \frac{1}{-5}$
        $0 - \frac{1}{20 - d} = -\frac{1}{5}$
        $\frac{1}{20 - d} = \frac{1}{5}$
        $20 - d = 5 \implies d = 15	ext{ cm}$.

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