A wire mesh consisting of very small squares | vedclass

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

Question

(A) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Given,the object distance $u = -8 \, cm$ (using sign convention) and the fo

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Given,the object distance $u = -8 \, cm$ (using sign convention) and the focal length $f = +10 \, cm$ for a converging lens.Substituting these values into the lens formula:$\frac{1}{v} - \frac{1}{-8} = \frac{1}{10}$$\frac{1}{v} + \frac{1}{8} = \frac{1}{10}$$\frac{1}{v} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = -\frac{1}{40}$Thus,$v = -40 \, cm$.The magnification $m$ produced by the lens is given by $m = \frac{v}{u}$.$m = \frac{-40}{-8} = 5$.Therefore,the magnification produced by the lens is $5$.

$A$ wire mesh consisting of very small squares is viewed at a distance of $8 \, cm$ through a magnifying converging lens of focal length $10 \, cm$,kept close to the eye. The magnification produced by the lens is

Similar Questions

What is a simple microscope?

The magnification power of a compound microscope is given in terms of the magnification of the objective $m_0$ and the magnification power of the eyepiece $m_E$. The total magnification is:

The focal lengths of the objective and eye-lens of a microscope are $1 \,cm$ and $5 \,cm$ respectively. If the magnifying power for the relaxed eye is $45$,then the length of the tube is $....... \,cm$.

In a compound microscope,the magnification will be large if the focal length of the eyepiece 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module