Question

In a compound microscope the objective and eye-piece have focal lengths of $\mathrm{0..95}$cm and $5cm$ respectively, and are kept at a distance of $20cm$. The last image is formed at a distance of $25cm$ from eye-piece. What is the total magnification of the microscope:-

• A. $95$
• B. $-94$
• C. $\frac{94}{6}$
• D. None of these

Answer 1

Answer: (B)

$-94$

As final image is at $25cm$ in front of eye piece,

$\frac{-1}{25}-\frac{1}{u_e}=\frac{1}{5}$

That is,

$u_e=\frac{-25}{6}$

And also,

$m_e=\frac{v_e}{u_e}=\frac{-25}{-25/6}=6$

For objective,

$v-L-u_e$

$=20-(\frac{25}{6}=\frac{95}{6}$

If object is at a distance $u$ from the objective,

$\frac{6}{95}=\frac{1}{u}=\frac{1}{0.95}$

That is,

$u=\frac{95}{94}cm$

That is, object is at a distance $\left(\frac{95}{94}\right)cm$ in front of field lens.

Also,

$m=\frac{v}{u}=\frac{\left(95/6\right)}{(-95/94)}=\left[\frac{94}{6}\right]$

So, total magnification,

$M=m\times m_e=-\left[\frac{94}{6}\right]\times 6=-94$

Final image is inverted, virtual and $94$ times that of object.