Question

A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?

Answer 1

For the compound microscope, we have:
Focal length of the objective, f_o = 1.0 cm
Focal length of the eyepiece, f_e = 5 cm
Distance of the object from the objective, u₀ = 0.5 cm
Distance of the image from the eyepiece, v_e = 30 cm
The lens formula for the objective lens is given by
$$\begin{gathered} \frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0} \\ \Rightarrow \frac{1}{v_0}+\frac{1}{0.5}=\frac{1}{1} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{v_0}=1-\frac{10}{5}=-1 \\ \Rightarrow v_0=-1\mathrm{cm} \end{gathered}$$
The objective will form a virtual image at the side same as that of the object at a distance of 1 cm from the objective lens. The image formed by the objective will act as a virtual object for the eyepiece.
$$\begin{gathered} \mathrm{The}\mathrm{lens}\mathrm{formula}\mathrm{for}\mathrm{the}\mathrm{eyepiece}\mathrm{is}\mathrm{given}\mathrm{by} \\ \frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{30}-\frac{1}{u_e}=\frac{1}{5} \\ \Rightarrow -\frac{1}{u_e}=\frac{1}{5}-\frac{1}{30}=\frac{6-1}{30}=\frac{1}{6} \end{gathered}$$
⇒ u_e = − 6 cm
∴ Separation between the objective and the eyepiece = 6 − 1 = 5 cm