Question

Find the nature, position and magnification of the image
formed by a convex lens of focal length 10cm, If the
object is placed at a distance of
a) 15cm b) 8cm

Answer 1

Answer:

the nature is real inverted and diminished

Answer 2

Answer:

(a):

Nature: Real and inverted.

Position = 30 cm behind the lens.

Magnification = -2.

(b):

Nature: Virtual and erect.

Position = 40 cm in front of the lens.

Magnification = 5.

Explanation:

Sign convention used:

The distances are taken as positive if measured along the right side or above the principal axis of the lens and the distances are taken as negative if measured along the left side of below the principal axis of the lens.

Given:

• Focal length of the convex lens, $f=10\ cm$.

• Object distance is $u$.

• Image distance is $v$.

(a): $u = -15\ cm$.

Using lens equation,

$\dfrac 1f = \dfrac 1v-\dfrac 1u\\\dfrac 1v = \dfrac 1f+\dfrac 1u$

$\dfrac 1v = \dfrac {1}{10}+\dfrac {1}{-15}= \dfrac {1}{10}-\dfrac {1}{15}=\dfrac{15-10}{10\times 15}=\dfrac{5}{150} = \dfrac{1}{30}\\\Rightarrow v = +30\ cm.$

Positive value of $v$ indicates that the image is formed behind the lens and therefore it is real in nature.

The magnification is given as

$m=\dfrac{v}{u}=\dfrac{30}{-15}=-2.$

Negative value of magnification indicates the image formed is inverted.

(b): $u = -8\ cm$.

Using lens equation,

$\dfrac 1f = \dfrac 1v-\dfrac 1u\\\dfrac 1v = \dfrac 1f+\dfrac 1u$

$\dfrac 1v = \dfrac {1}{10}+\dfrac {1}{-8}= \dfrac {1}{10}-\dfrac {1}{8}=\dfrac{8-10}{10\times 8}=\dfrac{-2}{80} = \dfrac{-1}{40}\\\Rightarrow v = -40\ cm.$

Negative value of $v$ indicates that the image is formed in front of the lens and therefore it is virtual in nature.

The magnification is given as

$m=\dfrac{v}{u}=\dfrac{-40}{-8}=5.$

Positive value of magnification indicates the image formed is upright(erect).