# (a) Find the nature, position and magnification of the images formed by a convex lens of focal length 0.20 m if the object is placed at a distance of: (i) 0.50 m (ii) 0.25 m (iii) 0.15 m(b) Which of the above cases represents the use of convex lens in a film projector, in a camera, and as a magnifying glass?

(a) (i)Given:

Object distance, $u$ = $-$0.50 m = $-$50 cm     (negative sign shows that the object is placed on the left side of the lens)

Focal length, $f$ = 0.20 m = 20 cm

To find: Nature, position, $(v)$ and magnification $(m)$ of the image.

Solution:

According to the lens formula, we know that:

$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$

Substituting the given values in the formula we get-

$\frac {1}{v}-\frac {1}{(-50)}=\frac {1}{20}$

$\frac {1}{v}+\frac {1}{50}=\frac {1}{20}$

$\frac {1}{v}=\frac {1}{20}-\frac {1}{50}$

$\frac {1}{v}=\frac {5-2}{100}$

$\frac {1}{v}=\frac {3}{100}$

$v=\frac {100}{3}$

$v=+33.3cm$

Thus, the image $v$ is formed at a distance of 33.3 cm from the convex lens, and the positive $(+)$ sign for image distance implies that the image is placed on the right side of the convex lens.

Now,

According to the magnification formula, we know that:

$m=\frac {v}{u}$

Substituting the given values in the formula we get-

$m=\frac {33.3}{-50}$

$m=-0.66$

Thus, the magnification $m$ of the image is 0.66, and the negative sign $(-)$ implies that the image is real and inverted.

Hence, the nature of the image is real and inverted, the position of the mirror is on the right side of the lens, and the magnification is 0.66.

(ii) Given:

Object distance, $u$ = $-$0.25 m = $-$25 cm     (negative sign shows that the object is placed on the left side of the lens)

Focal length, $f$ = 0.20 m = 20 cm

To find: Nature, position, $(v)$ and magnification $(m)$ of the image.

Solution:

According to the lens formula, we know that:

$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$

Substituting the given values in the formula we get-

$\frac {1}{v}-\frac {1}{(-25)}=\frac {1}{20}$

$\frac {1}{v}+\frac {1}{25}=\frac {1}{20}$

$\frac {1}{v}=\frac {1}{20}-\frac {1}{25}$

$\frac {1}{v}=\frac {5-4}{100}$

$\frac {1}{v}=\frac {1}{100}$

$v=+100cm$

Thus, the image $v$ is formed at a distance of 100 cm from the convex lens, and the positive $(+)$ sign for image distance implies that the image is placed on the right side of the convex lens.

Now,

According to the magnification formula, we know that:

$m=\frac {v}{u}$

Substituting the given values in the formula we get-

$m=\frac {100}{-25}$

$m=-4$

Thus, the magnification $m$ of the image is 4, and the negative sign $(-)$ implies that the image is real and inverted.

Hence, the nature of the image is real and inverted, the position of the mirror is on the right side of the lens, and the magnification is 4.

(iii) Given:

Object distance, $u$ = $-$0.15 m = $-$15 cm     (negative sign shows that the object is placed on the left side of the lens)

Focal length, $f$ = 0.20 m = 20 cm

To find: Nature, position, $(v)$ and magnification $(m)$ of the image.

Solution:

According to the lens formula, we know that:

$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$

Substituting the given values in the formula we get-

$\frac {1}{v}-\frac {1}{(-15)}=\frac {1}{20}$

$\frac {1}{v}+\frac {1}{15}=\frac {1}{20}$

$\frac {1}{v}=\frac {1}{20}-\frac {1}{15}$

$\frac {1}{v}=\frac {3-4}{60}$

$\frac {1}{v}=-\frac {1}{60}$

$v=-60cm$

Thus, the image $v$ is formed at a distance of 60 cm from the convex lens, and the positive $(-)$ sign for image distance implies that the image is placed on the left side of the convex lens.

Now,

According to the magnification formula, we know that:

$m=\frac {v}{u}$

Substituting the given values in the formula we get-

$m=\frac {-60}{-15}$

$m=+4$

Thus, the magnification $m$ of the image is 4, and the positive sign $(+)$ implies that the image is virtual and erect.

Hence, the nature of the image is virtual and erectthe position of the mirror is on the left side of the lens, and the magnification is 4.

(b) The above cases that represent the use of convex lens in a film projector, in a camera, and as a magnifying glass are:-

Case (ii) can be used for a film projector.

Case (i) can be used for a camera.

Case (iii) can be used for a magnifying glass.

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Updated on: 10-Oct-2022

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