Question

The length of the tube of a microscope is $10cm$.The lengths of the objective and eye lenses are $0.5cm$ and $1cm$ respectively. The magnifying power of the microscope when the images at far point is about

• A. $5$
• B. $23$
• C. $166$
• D. $500$

Answer 1

The correct option is D $500$

Given,
Length of tube microscope, $L=10cm$

Focal length of objective lens, $f_o=0.5cm$

Focal length of eye lens, $f_e=1cm$

The image is formed at the far point. Hence, using the formula for normal adjustment, we have
$m=\frac{v_o}{u_o}\frac{D}{f_e}$
where, $D$ is distance of distinct vision and $D=25cm$
It can be assumed that, $u_o=f_o$ because object is close to objective lens' first focus and that $v_o\approx L$ because $u_e<<v$. Thus, the formula becomes,

$\Rightarrow m=\frac{L}{f_o}\frac{D}{f_e}$
$\Rightarrow m=\frac{10}{0.5}\times \frac{25}{1}=\frac{250}{0.5}=500$
Hence, the correct answer is OPTION D.