A microscope has an objective of focal length | vedclass

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

Question

(C) For a relaxed eye,the final image is formed at infinity. The distance between the objective and the eyepiece is given by $L = v_0 + f_e$,where $v_

Vedclass · Accepted Answer

$140$

Vedclass · Answer

(C) For a relaxed eye,the final image is formed at infinity. The distance between the objective and the eyepiece is given by $L = v_0 + f_e$,where $v_0$ is the image distance from the objective and $f_e$ is the focal length of the eyepiece.Given $L = 25\, cm$ and $f_e = 2.5\, cm$,we have $v_0 = L - f_e = 25 - 2.5 = 22.5\, cm$.Using the lens formula for the objective lens,$\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$,where $f_0 = 1.5\, cm$:$\frac{1}{22.5} - \frac{1}{u_0} = \frac{1}{1.5} \implies \frac{1}{u_0} = \frac{1}{22.5} - \frac{1}{1.5} = \frac{1 - 15}{22.5} = -\frac{14}{22.5}$.Thus,$|u_0| = \frac{22.5}{14} \approx 1.607\, cm$.The magnification for a relaxed eye is given by $m = m_0 	imes m_e = (\frac{v_0}{|u_0|}) 	imes (\frac{D}{f_e})$,where $D = 25\, cm$ is the least distance of distinct vision.$m = (\frac{22.5}{1.607}) 	imes (\frac{25}{2.5}) = 14 	imes 10 = 140$.

$A$ 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?

Similar Questions

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.

$A$ card sheet divided into squares each of size $1\,mm^2$ is being viewed at a distance of $9\,cm$ through a magnifying glass (a converging lens of focal length $9\,cm$) held close to the eye. What is the angular magnification (magnifying power) of the lens?

$A$ microscope has an objective of focal length $1 \ cm$ and an eye-piece of focal length $6 \ cm$. If the tube length is $30 \ cm$ and the image is formed at the least distance of distinct vision,what is the magnification produced by the microscope? Take $D = 25 \ cm$.

The magnifying power of a simple microscope is $6$. The focal length of its lens in metres will be,if the least distance of distinct vision is $25\,cm$.

The angular magnification of a simple microscope can be increased by increasing

Vedclass Test Series

Exam Paper Generator

Online Exam Module