The length of the compound microscope is 14 , | vedclass

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

Question

(A) For a compound microscope,the length of the tube $L$ for a relaxed eye (final image at infinity) is given by $L = v_o + f_e$,where $v_o$ is the im

Vedclass · Accepted Answer

$1.8$

Vedclass · Answer

(A) For a compound microscope,the length of the tube $L$ for a relaxed eye (final image at infinity) is given by $L = v_o + f_e$,where $v_o$ is the image distance of the objective lens and $f_e$ is the focal length of the eyepiece.Given $L = 14 \, cm$ and $f_e = 5 \, cm$,we have $14 = v_o + 5$,which gives $v_o = 9 \, cm$.The magnifying power $m$ for a relaxed eye is given by $m = \frac{v_o}{u_o} 	imes \frac{D}{f_e}$,where $D$ is the least distance of distinct vision $(25 \, cm)$.Substituting the values: $25 = \frac{9}{u_o} 	imes \frac{25}{5}$.$25 = \frac{9}{u_o} 	imes 5$.$25 = \frac{45}{u_o}$.$u_o = \frac{45}{25} = 1.8 \, cm$.

The length of the compound microscope is $14 \, cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $5 \, cm$,then the object distance for the objective lens will be.......$ cm$.

Similar Questions

The least distance of distinct vision is $25 \ cm$. Find the magnifying power of a simple microscope of focal length $5 \ cm$ if the final image is formed at the least distance of distinct vision.

$A$ simple magnifying lens is used in such a way that an image is formed at $25 \, cm$ away from the eye. In order to have $10$ times magnification,the focal length of the lens should be

The image formed by the objective lens of a compound microscope is:

To obtain a magnified image at the distance of distinct vision $(D)$ using a simple microscope,the object should be placed:

Vedclass Test Series

Exam Paper Generator

Online Exam Module