A converging beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at

Question

A converging beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at a point $15	ext{ cm}$ from the lens on the opposite side. If the lens is removed the point where the rays meet will move $5	ext{ cm}$ closer to the lens. The focal length of the lens is:

Accepted Answer

1.  **Image Distance ($v$):** With the diverging lens in place, the rays intersect at a distance of $15	ext{ cm}$ on the opposite side. This is the position of the real image formed by the lens. Therefore, $v = +15	ext{ cm}$.
2.  **Object Distance ($u$):** If the lens is removed, the rays would have met $5	ext{ cm}$ closer to the lens. This means they would have intersected at a distance of $15	ext{ cm} - 5	ext{ cm} = 10	ext{ cm}$ from the lens's position. This point acts as a **virtual object** for the diverging lens. Therefore, the object distance $u = +10	ext{ cm}$.
3.  **Lens Formula:** Using the thin lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
    Substituting the values with their proper sign convention:
    $\frac{1}{+15} - \frac{1}{+10} = \frac{1}{f}$
    $\frac{1}{f} = \frac{2 - 3}{30} = -\frac{1}{30}$
    $f = -30	ext{ cm}$.
4.  **Conclusion:** The focal length of the diverging lens is $-30	ext{ cm}$. The negative sign confirms it is a concave (diverging) lens.

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