Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A convex lens A of focal length $20\text{ cm}$ and a concave lens B of focal length $5\text{ cm}$ 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:

Step-by-Step Solution

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 $v_A = f_A = +20\text{ cm}$ from its optical center.
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).
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: $d = f_A + f_B$ .
Calculation: $f_A = +20\text{ cm}$ (Convex) $f_B = -5\text{ cm}$ (Concave)

$d = 20 + (-5) = 15\text{ cm}$ .

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 $20\text{ cm}$ 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: $\frac{1}{\infty} - \frac{1}{20 - d} = \frac{1}{-5}$ $0 - \frac{1}{20 - d} = -\frac{1}{5}$ $\frac{1}{20 - d} = \frac{1}{5}$ $20 - d = 5 \implies d = 15\text{ cm}$ .

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