Q112 / 120 · VITEEE 2015 · Physics

A compound microscope having magnifying power 35 with its eye-piece of focal length 10 $\mathrm{cm}$. Assume that the final image is at least distance of distinct vision then the magnification produced by the objective is

  • A
    $-4$
  • B
    5
  • C
    10
  • Correct answer
    D
    $-10$

Solution

For a compound microscope, magnifying power $\mathrm{MP}=\mathrm{m}_{\mathrm{e}} \times \mathrm{m}_{0}$ When the final image is at least distance of distance vision then $\begin{aligned} & \mathrm{M}_{\mathrm{e}}=1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}} \\ \therefore \quad & \mathrm{MP}=\mathrm{m}_{0}\left[1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right] \\ \Rightarrow \quad &-35=\mathrm{m}_{0}\left[1+\frac{25}{10}\right] \\ \Rightarrow \quad &-35=\mathrm{m}_{0} \times 35 \\ \Rightarrow \quad & \mathrm{m}_{0}=-10 \end{aligned}$ The negative sign shows that the image formed by objective is inverted.

Hence, the correct answer is option: 'D'.

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