Magnifying power of a simple microscope is (w | vedclass

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

Question

(B) For a simple microscope,the magnifying power $M$ is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by

Vedclass · Accepted Answer

$1 + D/f$

Vedclass · Answer

(B) For a simple microscope,the magnifying power $M$ is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the near point.When the final image is formed at the near point $D = 25 \ cm$,we use the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Here,$v = -D$ and $u = -u$. Substituting these values,we get: $\frac{1}{-D} - \frac{1}{-u} = \frac{1}{f}$.Rearranging the terms: $\frac{1}{u} = \frac{1}{f} + \frac{1}{D}$.Multiplying both sides by $D$,we get: $\frac{D}{u} = \frac{D}{f} + 1$.Since the magnifying power $M = \frac{D}{u}$,we have $M = 1 + \frac{D}{f}$.

Magnifying power of a simple microscope is (when final image is formed at $D = 25 \ cm$ from eye)

Similar Questions

In a compound microscope,the intermediate image is

The intermediate image formed by an objective lens of a compound microscope is

Obtain the magnification for the image formed at infinity for a simple microscope.

$A$ microscope has an objective of focal length $1 \ cm$ and an eyepiece of focal length $6 \ cm$. If the tube length is $30 \ cm$ and the image is formed at the least distance of distinct vision,what is the magnification produced by the microscope? Take $D = 25 \ cm$.

In a compound microscope,the focal length and the aperture of the objective lens used are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module