(a) Two lenses A and B have power of (i) +2D and (ii) −4D respectively. What is the nature and focal length of each lens?(b) An object is placed at a distance of 100 cm from each of the above lenses A and B. Calculate (i) image distance, and (ii) magnification, in each of the two cases.

 Power of the lens is given by-

$Power\ (P)=\frac {1}{fcoal\ length\ (f)}$

or, $f=\frac {1}{P}$

Therefore,

(i) The focal length of the lens A, $(f_A)=\frac {1}{2}=+0.5m=+50cm$

(ii) The focal length of the lens B, $(f_B)=\frac {1}{-4}=-0.25m=-25cm$

Thus, the focal length of lens A is 50 cm, and the positive sign $(+)$ implies that the nature of the lens is convex, and

The focal length of lens B is 25 cm, and the negative sign $(-)$ implies that the nature of the lens is concave.

(b) For lens A

Given: 

Object distance, $u$ = $-$100 cm (object distance is always taken negative)

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

Solution:

From the lens formula, we know that-

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

Substituting the given value we get-

$\frac {1}{v}-\frac {1}{(-100)}=\frac {1}{50}$

$\frac {1}{v}+\frac {1}{100}=\frac {1}{50}$

$\frac {1}{v}=\frac {1}{50}-\frac {1}{100}$

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

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

$v=+100cm$

Thus, the image distance, $v$ is 100 cm behind the lens.

Now, 

From the magnification formula, we know that-

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

Substituting the given value we get-

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

$m=-1$

Thus, the magnification, $m$ of the lens is 1.

For lens B

Given: 

Object distance, $u$ = $-$100 cm (object distance is always taken negative)

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

Solution:

From the lens formula, we know that-

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

Substituting the given value we get-

$\frac {1}{v}-\frac {1}{(-100)}=\frac {1}{(-25)}$

$\frac {1}{v}+\frac {1}{100}=-\frac {1}{25}$

$\frac {1}{v}=-\frac {1}{25}-\frac {1}{100}$

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

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

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

$v=-20cm$

Thus, the image distance, $v$ is 20 cm in front of the lens.

Now, 

From the magnification formula, we know that-

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

Substituting the given value we get-

$m=\frac {-20}{-100}$

$m=\frac {1}{5}$

$m=-0.2$

Thus, the magnification, $m$ of the lens is 0.2.