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A converging beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at a point 15 cm15\text{ cm} from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm5\text{ cm} closer to the lens. The focal length of the lens is:

A

−10 cm-10\text{ cm}

B

20 cm20\text{ cm}

C

−30 cm-30\text{ cm}

D

5 cm5\text{ cm}

Step-by-Step Solution

  1. Image Distance (vv): With the diverging lens in place, the rays intersect at a distance of 15 cm15\text{ cm} on the opposite side. This is the position of the real image formed by the lens. Therefore, v=+15 cmv = +15\text{ cm}.
  2. Object Distance (uu): If the lens is removed, the rays would have met 5 cm5\text{ cm} closer to the lens. This means they would have intersected at a distance of 15 cm−5 cm=10 cm15\text{ cm} - 5\text{ cm} = 10\text{ cm} from the lens's position. This point acts as a virtual object for the diverging lens. Therefore, the object distance u=+10 cmu = +10\text{ cm}.
  3. Lens Formula: Using the thin lens formula: 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} Substituting the values with their proper sign convention: 1+15−1+10=1f\frac{1}{+15} - \frac{1}{+10} = \frac{1}{f} 1f=2−330=−130\frac{1}{f} = \frac{2 - 3}{30} = -\frac{1}{30} f=−30 cmf = -30\text{ cm}.
  4. Conclusion: The focal length of the diverging lens is −30 cm-30\text{ cm}. The negative sign confirms it is a concave (diverging) lens.
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