JEE PYQ: Ray Optics Question 41
Question 41 - 2019 (10 Apr Shift 2)
The graph shows how the magnification $m$ produced by a thin lens varies with image distance $v$. What is the focal length of the lens used?
(1) $\frac{b^2}{ac}$
(2) $\frac{b^2 c}{a}$
(3) $\frac{a}{c}$
(4) $\frac{b}{c}$
Show Answer
Answer: (4)
Solution
From the equation of line $m = k_1 v + k_2$ ($\because y = mx + c$) $\Rightarrow \frac{v}{u} = k_1 v + k_2$ $\left(\because m = \frac{v}{u}\right)$ $\Rightarrow \frac{1}{u} = k_1 + \frac{k_2}{v}$ (Dividing both sides by $v$) $\Rightarrow \frac{k_2}{v} = \frac{1}{u} - k_1$ Comparing with lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, we get $k_1 = \frac{1}{f}$ and $k_2 = 1$ $\therefore f = \frac{1}{\text{slope of } m - v \text{ graph}} = \frac{b}{c}$
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