The focal length of the objective lens and th | vedclass

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

Question

(A) Given: Focal length of objective $f_o = 2 \, cm$,focal length of eyepiece $f_e = 5 \, cm$,tube length $L = 20 \, cm$,and final image distance $v_e

Vedclass · Accepted Answer

$2.3$

Vedclass · Answer

(A) Given: Focal length of objective $f_o = 2 \, cm$,focal length of eyepiece $f_e = 5 \, cm$,tube length $L = 20 \, cm$,and final image distance $v_e = -25 \, cm$.First,find the object distance for the eyepiece $(u_e)$ using the lens formula $\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}$:$\frac{1}{5} = \frac{1}{-25} - \frac{1}{u_e} \implies \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25} \implies u_e = -\frac{25}{6} \, cm$.The magnitude of the object distance for the eyepiece is $|u_e| = \frac{25}{6} \approx 4.17 \, cm$.The distance between the lenses is $L = |v_o| + |u_e|$,where $v_o$ is the image distance from the objective lens:$20 = v_o + 4.17 \implies v_o = 15.83 \, cm$.Now,use the lens formula for the objective lens to find the object distance $u_o$:$\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \implies \frac{1}{2} = \frac{1}{15.83} - \frac{1}{u_o}$.$\frac{1}{u_o} = \frac{1}{15.83} - \frac{1}{2} = \frac{2 - 15.83}{31.66} = -\frac{13.83}{31.66}$.$u_o = -\frac{31.66}{13.83} \approx -2.29 \, cm$.The distance of the object from the objective lens is approximately $2.3 \, cm$.

The focal length of the objective lens and the eyepiece of a microscope are $2 \, cm$ and $5 \, cm$ respectively,and the distance between them is $20 \, cm$. Find the distance of the object from the objective lens when the final image is seen by the eye at $25 \, cm$ from the eyepiece.

Similar Questions

The objective of a compound microscope is essentially

In a compound microscope,the focal length of the objective and eyepiece lens is $4\,cm$ and $8\,cm$ respectively. If the length of the tube is $30\,cm$,then the magnifying power for normal adjustment is:

The objective lens of a compound microscope produces a magnification of $10$. In order to get an overall magnification of $100$ when the image is formed at $25 \, cm$ from the eye,the focal length of the eye lens should be:

$A$ compound microscope consists of an eyepiece of focal length $6.25 \, cm$ and an objective lens of focal length $2.0 \, cm$,separated by a distance of $15 \, cm$. What is the magnifying power when the final image is formed at infinity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module