(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.
(a) (i) Power of lens A = $+$2D
(ii) Power of lens B = $-$4D
To find: Nature and focal length of the lens.
Solution:
 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.
Related Articles
- Two lenses A and B have focal lengths of +20 cm and, −10 cm, respectively.(a) What is the nature of lens A and lens B?(b) What is the power of lens A and lens B?(c) What is the power of combination if lenses A and B are held close together?
- What is meant by power of a lens? Define its SI unit. You have two lenses A and B of focal lengths +10 cm and –10 cm, respectively. State the nature and power of each lens. Which of the two lenses will form a virtual and magnified image of an object placed 8 cm from the lens? Draw a ray diagram to justify your answer.
- An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm.(i) What is the nature of image?(ii) What is the position of image?
- 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.
- An object of height 4.25 mm is placed at a distance of 10 cm from a convex lens of power +5D. Find (i) focal length of the lens, and (ii) size of the image.
- A convex lens of power 5 D and a concave lens of power 7.5 D are placed in contact with each other. What is the :(a) power of this combination of lenses?(b) focal length of this combination of lenses?
- An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Fine the position and nature of the image.
- The power of a combination of two lenses X and Y is 5 D. If the focal length of lens X be 15 cm :(a) calculate the focal length of lens Y.(b) state the nature of lens Y.
- When an object is placed 10 cm in front of lens A, the image is real, inverted, magnified and formed at a great distance. When the same object is placed 10 cm in front of lens B, the image formed is real, inverted and same size as the object.(a) What is the focal length of lens A?(b) What is the focal length of lens B?(c) What is the nature of lens A?(d) What is the nature of lens B?
- (a) A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the position, nature and size of the image formed.(b) Draw a labelled ray diagram showing object distance, image distance and focal length in the above case.
- (a) An object 2 cm tall stands on the principal axis of a converging lens of focal length 8 cm. Find the position, nature and size of the image formed if the object is:(i) 12 cm from the lens(ii) 6 cm from the lens(b) State one practical application each of the use of such a lens with the object in position (i) and (ii).
- An object is placed at a distance of 10 cm from a convex mirror of focal length 5 cm.(i) Draw a ray-diagram showing the formation image(ii) State two characteristics of the image formed(iii) Calculate the distance of the image from mirror.
- (a) What do you understand by the power of a lens? Name one factor on which the power of a lens depends.(b) What is the unit of power of a lens? Define the unit of power of a lens.(c) A combination of lenses for a camera contains two converging lenses of focal lengths 20 cm and 40 cm and a perging lens of focal length 50 cm. Find the power and focal length of the combination.
- An object is placed at a distance of 10 cm from a convex mirror of focal length 8 cm. Find the position and nature of the image.
- An object is placed at a distance of 6 cm from a convex mirror of focal length 12 cm. Find the position and nature of the image.
Kickstart Your Career
Get certified by completing the course
Get Started