Question

A compound microscope has a magnifying power 30. The focal length of its eye-piece is 5 cm. Assuming the final image to be at the least distance of distinct vision. The magnification produced by the objective will be

• A. +5
• B. – 5
• C. +6
• D. – 6

Answer 1

Answer: (B)

– 5

Answer 2

Magnification produced by compound microscope $m = m_{o} \times m_{e}$

where m_o = ? and $m_{e} = \left( 1 + \frac{D}{f_{e}} \right) = 1 + \frac{25}{5} = 6$

⇒ $30 = - m_{o} \times 6$ ⇒ $m_{o} = - 5$.