Question

In a compound microscope, the linear magnification due to the objective is \[m_o = \frac{h'}{h}\]. Which relation is used to obtain the result \[m_o = \frac{L}{f_o}\]?

Options

• \[\tan\beta = \frac{h}{f_o} = \frac{h'}{L}\]

• \[\tan\beta = \frac{h}{D} = \frac{h'}{f_e}\]

• \[\tan\beta = \frac{L}{f_o} = \frac{h'}{h}\]

• \[\tan\beta = \frac{h'}{f_o} = \frac{h}{L}\]

Answer 1

The linear magnification due to the objective is \[m_o = \frac{h'}{h} = \frac{L}{f_o}\]. This result uses the relation \[\tan\beta = \frac{h}{f_o} = \frac{h'}{L}\], where \[L\] is the tube length and \[f_o\] is the focal length of the objective.