A compound microscope consists of an objectiv | vedclass

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

Question

(A) For the eyepiece,the final image is at the least distance of distinct vision,so $v_e = -25 \ cm$. Given $f_e = 6.25 \ cm$. Using the lens formula

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) For the eyepiece,the final image is at the least distance of distinct vision,so $v_e = -25 \ cm$. Given $f_e = 6.25 \ cm$. Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:$\frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} = \frac{1}{-25} - \frac{1}{6.25} = \frac{-1 - 4}{25} = -\frac{5}{25} = -\frac{1}{5} \ cm^{-1}$.So,$u_e = -5 \ cm$. The distance between the lenses is $L = v_0 + |u_e| = 15 \ cm$.$v_0 + 5 = 15 \implies v_0 = 10 \ cm$.For the objective lens,$f_0 = 2 \ cm$ and $v_0 = 10 \ cm$. Using the lens formula $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$:$\frac{1}{u_0} = \frac{1}{v_0} - \frac{1}{f_0} = \frac{1}{10} - \frac{1}{2} = \frac{1 - 5}{10} = -\frac{4}{10} = -\frac{2}{5} \ cm^{-1}$.$u_0 = -2.5 \ cm$.Thus,the object should be placed $2.5 \ cm$ from the objective lens.

Similar Questions

The focal lengths of the objective and eyepiece of a compound microscope are $1 \, cm$ and $5 \, cm$ respectively. If the magnification is $45$,what is the length of the tube in $cm$?

$A$ 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.

In a compound microscope,the intermediate image is

The least distance of distinct vision is $25 \, cm$. The magnifying power of a simple microscope with a focal length of $5 \, cm$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module