Question

A microscope has an objective of focal length $1.5$ cm and an eyepiece of focal length $2.5cm$. If the distance between objective and eyepiece is $20.5cm$, what is the magnification produced when the final image is at infinite distance away?

• A. $100$
• B. $75$
• C. $110$
• D. $120$

Answer 1

Answer: (C)

$110$

$f_0=1.5cm$

$f_e=2.5cm$

$L=20.5cm$

$M_{0D}=\frac{v_{0}D}{v_0{f}_e}$

$L=v_0+f_e$

$20.5=v_0+2.5$

$v_0=18cm$

$\frac{\text{/}1}{\text{/}18}-\frac{1}{4}=\frac{+12}{18}-\frac{1}{18}$

$-\frac{1}{u}=\frac{+11}{18}$

$M=\frac{18}{18}\times 11\times \frac{25}{2.5}=110$