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 -160-of focal length - 9 cm- held close to the eye-In order to view the squares distinctly with the -160-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-c- Is the magnification in -a- equal to the magnifying power in -b-Explain-

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Standard XII

Physics

Magnification by Thin Lenses

A card sheet ...

Question

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?
(c) Is the magnification in (a) equal to the magnifying power in (b)?
Explain:

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Solution

(a) Area of each square , $A = 1 m m^{2}$
Object distance $u = - 9 c m$
Focal length of the converging lens , $f = 10 c m$
For image distance $v$ , the lens formula can be written as
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
By putting the values we get $v = - 90 c m$
Now magnification,
$m = \frac{u}{v} = \frac{- 60}{- 9} = 10$
$∴$ Area of each square in the virtual image = $10^{2} A = 10^{2} \times 1 = 100 m m^{2}$

(b) Magnifying power of the lens
$\frac{d}{| u |} = \frac{25}{9} = 2.8$

(c) The magnification in (a) is not same as the magnifying power in (b).
The ,magnification magnitude is
$(| \frac{v}{u} |)$ and the magnifying power is $(\frac{d}{| u |})$
The two quantities will be equal when the image is formed at the near point (25cm).

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A card sheet divided into squares each of size 1 mm² 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.

(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) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power? (b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case? Explain.

Q. 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.

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?

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