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The graph shows, how the inverse of magnification \mathrm{\frac{1}{m}} produce by a convex thin lens, with variation in object distance \mathrm{u} . What was the focal length of the lens used?

Option: 1

\mathrm{\frac{b}{c}}


Option: 2

\mathrm{\frac{b}{ca}}


Option: 3

\mathrm{\frac{bc}{a}}


Option: 4

\mathrm{\frac{c}{b}}


Answers (1)

best_answer

From,\ \frac{1}{v}-\frac{1}{-u}=\frac{1}{+f} \\ \begin{aligned} & \mathrm{v}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}} \text { or } \frac{\mathrm{v}}{\mathrm{u}}=\mathrm{m}=\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}} \\ & \Rightarrow \frac{1}{\mathrm{~m}}=\left(\frac{1}{\mathrm{f}}\right) \mathrm{u}-1 \end{aligned}
Comparing this with equation of straight line, \mathrm{\frac{1}{f}} is the slope of line, which is \mathrm{\frac{b}{c}} then f = \mathrm{\frac{c}{b}}
 

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jitender.kumar

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