When final image is formed at D.D.V from eye, the magnifying power of microscope is ( $f$ is the focal…

When final image is formed at D.D.V from eye, the magnifying power of microscope is ( $f$ is the focal length of the lens)
  1. $1+\frac{f}{D}$
  2. $1+\frac{D}{f}$
  3. $\frac{D}{f}$
  4. $1-\frac{D}{f}$

Solution

Consider the diagram below: Given, $v=(-D)$, Using mirror formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ $\begin{aligned} & \Rightarrow \frac{1}{(-D)}-\frac{1}{u}=\frac{1}{f} \\ & \Rightarrow \frac{1}{u}=-\left(\frac{f+D}{f D}\right)\end{aligned}$ Now, magnification can be easily calculated as: $\therefore m=\frac{v}{u}=\frac{(-D)(f+D)}{(-f D)}=\left(1+\frac{D}{f}\right)$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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