Question 187 of 295

For a concave lens of focal length $f$, the relation between object and image distance $u$ and $v,$ respectively, from its pole can best be represented by ($u=v$ is the reference line):

ABCD

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Answer: (A)

From the lens formula, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
$\Rightarrow v=f-\dfrac{{f}^{2}}{u+f}$
now, if we plot a graph between $u\text{and}v$

When $u\rightarrow \infty ;v=f-\dfrac{{f}^{2}}{\infty }=f$
and for $u\rightarrow 0;v=0$
Given that $u=v$is a reference line, therefore, $\dfrac{dv}{du}=0+\dfrac{{f}^{2}}{{\left(u+f\right)}^{2}}$
${\dfrac{dv}{du}}_{u=0}=0+\dfrac{{f}^{2}}{{\left(0+f\right)}^{2}}=1$
The equation of tangent of the curve at the origin, $v-0=1\left(u-0\right)\Rightarrow v=u$

JEE Main Physics practice question on Ray Optics (Lens Formula, Refraction of light from curved surface), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.