A microscope consists of an objective of foca | vedclass

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

Question

(B) For the eyepiece, the image distance $V_e = -25 \,cm$ and focal length $f_e = 5 \,cm$. Using the lens formula $\frac{1}{V_e} - \frac{1}{u_e} = \fr

Vedclass · Accepted Answer

$2.7$

Vedclass · Answer

(B) For the eyepiece, the image distance $V_e = -25 \,cm$ and focal length $f_e = 5 \,cm$. Using the lens formula $\frac{1}{V_e} - \frac{1}{u_e} = \frac{1}{f_e}$:$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5} \Rightarrow \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25} \Rightarrow u_e = -\frac{25}{6} \,cm$.The distance of the intermediate image from the objective is $v_0 = L - |u_e| = 10.5 - \frac{25}{6} = \frac{63 - 25}{6} = \frac{38}{6} \,cm$.For the objective lens, using the lens formula $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$ with $f_0 = 1.9 \,cm$:$\frac{6}{38} - \frac{1}{u_0} = \frac{1}{1.9} \Rightarrow \frac{1}{u_0} = \frac{6}{38} - \frac{10}{19} = \frac{6 - 20}{38} = -\frac{14}{38}$.$u_0 = -\frac{38}{14} \approx -2.71 \,cm$.Thus, the object should be placed at a distance of $2.7 \,cm$ from the objective.

Similar Questions

In a compound microscope,the focal length of the objective lens is $1.2 \, cm$ and the focal length of the eyepiece is $3.0 \, cm$. When the object is kept at $1.25 \, cm$ in front of the objective,the final image is formed at infinity. The magnifying power of the compound microscope is:

Which of the following statements are true in the context of a Compound Microscope?

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:

In order to increase the magnifying power of a compound microscope:

$A$ microscope has an objective of focal length $1.5\, cm$ and an eye-piece of focal length $2.5\, cm$. If the distance between objective and eye-piece is $25\, cm$,the approximate value of magnification produced for a relaxed eye is?

Vedclass Test Series

Exam Paper Generator

Online Exam Module