A compound microscope with an objective lens | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The magnifying power $(M)$ of a compound microscope when the final image is formed at the least distance of distinct vision $(D = 25 \, cm)$ is gi

Vedclass · Accepted Answer

$270$

Vedclass · Answer

(A) The magnifying power $(M)$ of a compound microscope when the final image is formed at the least distance of distinct vision $(D = 25 \, cm)$ is given by the formula:$M = \frac{L}{f_o} \left(1 + \frac{D}{f_e}ight)$Given:Tube length $(L)$ = $20 \, cm$Focal length of objective $(f_o)$ = $1 \, cm$Focal length of eyepiece $(f_e)$ = $2 \, cm$Least distance of distinct vision $(D)$ = $25 \, cm$Substituting the values into the formula:$M = \frac{20}{1} \left(1 + \frac{25}{2}ight)$$M = 20 	imes (1 + 12.5)$$M = 20 	imes 13.5$$M = 270$Therefore,the magnifying power of the microscope is $270$.

$A$ compound microscope with an objective lens of focal length $1 \, cm$ and an eyepiece of $2 \, cm$ focal length has a tube length of $20 \, cm$. Calculate the magnifying power of the microscope,if the final image is formed at the least distance of distinct vision.

Similar Questions

How should people wearing spectacles work with a microscope?

In a compound microscope,the intermediate image is generally .......

For a compound microscope,the focal lengths of the objective lens and the eye lens are ${f_o}$ and ${f_e}$ respectively. The microscope provides magnification when:

In a compound microscope,the objective of focal length ${f_o}$ and the eyepiece of focal length ${f_e}$ are placed at a distance $L$ such that $L$ equals:

In a compound microscope,the intermediate image is

Vedclass Test Series

Exam Paper Generator

Online Exam Module