A conversing beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at a point 15 ~cm from the lens on the opposite side. If the lens is removed the point where the rays meets

Correct answer: C. -30 ~cm Given u=10 ~cm, y=15 ~cm aligned & (1)/(f)=(1)/(y)-(1)/(u) & (1)/(f)=(1)/(15)-(1)/(10) & (1)/(f)=(10-15)/(150) & (1)/(f)=(-5)/(150) & f=-30 ~cm aligned

NEET 2011

Exam: NEET
Year: 2011
Subject: Physics
Chapter: Ray Optics

NEET 2011 · Physics · Ray Optics

A conversing beam of rays is incident on a diverging lens. Having...

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

15 \mathrm{~cm}

from the lens on the opposite side. If the lens is removed the point where the rays meets will move

5 \mathrm{~cm}

closer to the lens. The focal length of the lens is

Options

A $-10 \mathrm{~cm}$
B $20 \mathrm{~cm}$
C $-30 \mathrm{~cm}$
D $5 \mathrm{~cm}$

Correct answer

C. $-30 \mathrm{~cm}$

Step-by-step solution

Given

u=10 \mathrm{~cm}, y=15 \mathrm{~cm}

\begin{aligned}& \frac{1}{f}=\frac{1}{y}-\frac{1}{u}\\ & \frac{1}{f}=\frac{1}{15}-\frac{1}{10}\\ & \frac{1}{f}=\frac{10-15}{150}\\ & \frac{1}{f}=\frac{-5}{150}\\ & f=-30 \mathrm{~cm} \end{aligned}

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