Question

If the focal length of objective and eye lens are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is

• A. 150
• B. 200
• C. 250
• D. 400

Answer 1

Answer: (B)

200

Answer 2

$m_{\infty} = - \frac{v_{o}}{u_{o}} \times \frac{D}{f_{e}}$

From $\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}$

$\Rightarrow \frac{1}{( + 1.2)} = \frac{1}{v_{o}} - \frac{1}{( - 1.25)} \Rightarrow v_{o} = 30\mspace{6mu} cm$

$\therefore|m_{\infty}|\mspace{6mu} = \frac{30}{1.25} \times \frac{25}{3} = 200$