Question

# If the focal lengths of objective and eye-lens of a microscope are 1.2cm and 3cm respectively and the object is put 1.25cm away from the objective lens and the final image is formed at infinity, then magnifying power of the microscope is

A
150
B
200
C
250
D
400

Solution

## The correct option is B 200$$f_o = 1.2cm$$ $$u_e = f_e$$$$f_e = 3cm$$ $$m_\infty = \dfrac {v_0}{u_0} \dfrac {D}{f_e}$$$$u_0 = 1.25cm$$ $$L_\infty = r_0 + f_0$$$$\gamma_e = \infty$$$$M = ?$$$$\dfrac {1}{v_0} - \dfrac {1}{1.25} = \dfrac {-1}{1.2} \Rightarrow \dfrac {1}{v_0} = \dfrac {-1}{1.2} + \dfrac {1}{1.25} = \dfrac {-0.05}{1.2 \times 1.25} = \dfrac {-1}{30}$$$$v_0 = 30\ cm$$$$M_\infty = \dfrac {30}{1.25} \times \dfrac {2.5}{3} = 200$$PhysicsNCERTStandard XII

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