The total magnification produced by a compoun | vedclass

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

Question

(A) The total magnification $M$ 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

$4$

Vedclass · Answer

(A) The total magnification $M$ 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: $M = m_o 	imes m_e$, where $m_o$ is the magnification of the objective and $m_e$ is the magnification of the eyepiece.For an eyepiece acting as a simple magnifier, the magnification $m_e$ is given by: $m_e = (1 + D/f_e)$.Given $D = 25 \ cm$ and $f_e = 5 \ cm$, we calculate $m_e = (1 + 25/5) = (1 + 5) = 6$.Given the total magnification $M = 24$, we substitute the values into the formula: $24 = m_o 	imes 6$.Solving for $m_o$, we get $m_o = 24 / 6 = 4$.Therefore, the magnification produced by the objective is $4$.

The total magnification produced by a compound microscope is $24$ when the final image is formed at the least distance of distinct vision. If the focal length of the eyepiece is $5 \ cm$, the magnification produced by the objective is

Similar Questions

To improve the resolving power of a compound microscope, we should:

The objective of a compound microscope is essentially

The focal length of the objective lens of a compound microscope is

$A$ simple microscope is used to see an object first in blue light and then in red light. Due to the change from blue to red light, its magnifying power

The focal length of the objective and eye lens of a microscope are $4 \, cm$ and $8 \, cm$ respectively. If the least distance of distinct vision is $24 \, cm$ and the object distance is $4.5 \, cm$ from the objective lens, then the magnifying power of the microscope will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module