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.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 5
Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} + \frac{1}{8} = \frac{1}{10}$

$\frac{1}{v} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = - \frac{1}{40}$

$v = - 40 c m$
$m = \frac{v}{u} = \frac{40}{8} = 5$ .

Suggest Corrections

Similar questions

A card sheet divided into squares each of size

1 m m^{2}

is being viewed at a distance of

9 c m

through a magnifying glass (a converging lens of focal length

9 c m

) held close to the eye.

In order to view the squares distinctly with the maximum possible magnifying power.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

Explain:

Q. A card sheet divided into squares each of size

1 m m^{2}

is being viewed at a distance of

9 c m

through a magnifying glass (a converging lens of focal length

9 c m

) held close to the eye and the magnifying glass if the virtual image of each square is to have an area of

6.25 m m^{2}

. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

Q. A converging lens of focal length 6.25 cm is used as a magnifying glass. If near point of the observer is 25 cm from the eye and the lens is held close to the eye, then the magnification is :

Q. An object is kept at a distance of

20 c m

from a converging lens of focal length

10 c m

. What will be the magnification?

Answer the following questions:

(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?

(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?

(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?

(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?

(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

Join BYJU'S Learning Program

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.

The correct option is A 5Using 1v−1u=1f1v+18=1101v=110−18=4−540=−140v=−40cmm=vu=408=5.

The correct option is A 5
Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} + \frac{1}{8} = \frac{1}{10}$

$\frac{1}{v} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = - \frac{1}{40}$

$v = - 40 c m$
$m = \frac{v}{u} = \frac{40}{8} = 5$ .