Microscope and Telescope (Optical Instruments)
By Microscope
The magnification by a simple microscope of focal length $f$ is given by \begin{align} m=\frac{D}{f}, \end{align} where $D=25$ cm is the distant of distinct vision. This condition is called normal adjustment of the simple microscope.
A compound microscope has an objective of focal length $f_o$ and an eyepiece of focal length $f_e$. The image is formed at infinity in normal adjustment. The magnification of a compund microscope is given by \begin{align} m=\frac{v}{u}\frac{D}{f_e} \end{align}
The resolving power of a compund microscope is given by \begin{align} R=\frac{1}{\Delta d}=\frac{2\mu\sin\theta}{\lambda} \end{align}
Telescope
An astronomical telesocpe has an objective of focal length $f_o$ and an eyepiece of focal length $f_e$. In normal adjustment, the length of the telescope is given by \begin{align} L=f_o+f_e, \end{align} and its magnification is \begin{align} m=-\frac{f_o}{f_e} \end{align}
The resolving power of a telescope for light of wavelength $\lambda$ is given by \begin{align} R=\frac{1}{\Delta \theta}=\frac{1}{1.22\lambda}. \end{align}
Exercise
- A lady uses +1.5D glasses to have normal vision from 25 cm onwards. She uses a 20D lens as a simple microscope to see an object. Find the maximum magnifying power if she uses the microscope (a) together with her glasses (b) without her glasses. Do the answers suggest that an object can be more clearly seen through a microscope without using the correcting glasses?
