An object is placed 20 cm from (a) a converging lens, and (b) a perging lens, of focal length 15 cm. Calculate the image position and magnification in each case.

(a) Given:

A converging lens or convex lens.

Focal length of the convex lens, $f$ = $+$15 cm     (focal length of a convex lens is always taken positive)

Object distance from the lens, $u$ = $-$20 cm        (object distance is always taken negative, as it is placed on the left side of the lens)

To find: Image position or Image distance, $v$ and magnification, $m$.

Solution:

From the lens formula, we know that-

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

Substituting the given values in the formula, we get-

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

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

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

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

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

$v=+60cm$

Thus, the image is at a distance of 60 cm from the convex lens, and the positive sign implies that it is on the right side of it. Therefore, the image formed is real.

Now,

Fro magnification we know that-

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

Substituting the given values in the formula, we get-

$m=\frac {60}{-20}$

$m=-3$

Thus, the magnification, $m$ of the image is 3, which is greater than 1, so it implies that the image is magnified. The negative sign shows that the image is inverted.

(b) Given:

A diverging lens or concave lens.

Focal length of the convex lens, $f$ = $-$15 cm     (focal length of a concave lens is always taken negative)

Object distance from the lens, $u$ = $-$20 cm       (object distance is always taken negative, as it is placed on the left side of the lens)

To find: Image position or Image distance, $v$ and magnification, $m$.

Solution:

From the lens formula, we know that-

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

Substituting the given values in the formula, we get-

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

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

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

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

$\frac {1}{v}=-\frac {7}{60}$

$v=-8.57cm$

Thus, the image is at a distance of 8.57 cm from the concave lens, and the negative sign implies that it is on the left side of it. Therefore, the image formed is virtual. 

Now,

Fro magnification we know that-

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

Substituting the given values in the formula, we get-

$m=\frac {-8.57}{-20}$

$m=\frac {857}{20\times {100}}$

$m=\frac {857}{2\times {1000}}$

$m=\frac {428.5}{1000}$

$m=+0.42$

Thus, the magnification, $m$ of the image is 0.42, which is less than 1, so it implies that the image is diminished. The positive sign shows that the image is erect (upward).