Question

The length of the tube of a microscope is 10 cm. The focal lengths of
the objective and eye lenses are 0.5 cm and 1.0 cm. The magnifying
power of the microscope is about
(A) 5
(B) 23
(C) 166
(D) 500​

Answer 1

Explanation:

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Answer 2

Answer:

500

Explanation:

Given,  

Length of tube microscope, L=10 cm

Focal length of objective lens, fo  =0.5 cm

Focal length of eye lens, fe =1 cm

The image is formed at the far point. Hence, using the formula for normal adjustment, we have

m=$\frac{vo}{uo} \frac{D}{fo}$

where, D is distance of distinct vision and D=25 cm

It can be assumed that, uo  =fo

​because object is close to objective lens' first focus and that vo  ≈L because ue  <<v. Thus, the formula becomes,

⇒m=$\frac{L}{fo} \frac{D}{fe}$

⇒m=   $\frac{10}{0.5} into\frac{25}{1} =\frac{250}{0.5}=500$ =500

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