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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSMedium

Question

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 cm5 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: 1v1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} Substituting the values with their proper sign convention: 1+151+10=1f\frac{1}{+15} - \frac{1}{+10} = \frac{1}{f} 1f=2330=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.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from RAY OPTICS AND OPTICAL INSTRUMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSconvergingincidentdiverginghavingpassed

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