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

$A$ microscope has an objective of focal length $2 \ cm$,an eyepiece of focal length $4 \ cm$,and a tube length of $40 \ cm$. If the distance of distinct vision is $25 \ cm$,the magnification of the microscope is:

For which of the following colours will the magnifying power of a microscope be maximum?

What is a compound microscope? Explain its construction with the help of a figure.

For observing a cricket match,a binocular is preferred to a terrestrial telescope because

The magnifying power of a microscope with an objective of $5\, mm$ focal length is $400$. The length of its tube is $20\, cm$. Then the focal length of the eye-piece is.....$cm$

Vedclass Test Series

Exam Paper Generator

Online Exam Module