Question

a 10cm tall object is placed perpendicular to the principal Axis of a convex lens of focal length 30cm.the distance of the object from the line is 20cm.find the I) position ii) nature iii) size of the image formed​

Answer 1

Given :

• Height of object, $\sf{h_o}$ = 10 cm
• Focal length of convex lens, f = 30 cm
• Position of object, u = -20 cm

To find :

• Position of image, v = ?
• Size of image, $\sf{h_i}$ = ?
• Nature of image

Formulae required :

• Lens formula

$\red{\bigstar}\boxed{\sf{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}}$

• Formula for magnification produced by lens

$\red{\bigstar}\boxed{\sf{m=\dfrac{v}{u}=\dfrac{h_i}{h_o}}}$

[ where f is focal length of lens, v is position of image, u is position of object, $\sf{h_i}$ is height of image, $\sf{h_o}$ is height of object and m is magnification ]

Solution :

Calculating position of image

Using lens formula

$\longrightarrow\sf{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}$

$\longrightarrow\sf{\dfrac{1}{30}=\dfrac{1}{v}-\dfrac{1}{(-20)}}$

$\longrightarrow\sf{\dfrac{1}{30}=\dfrac{1}{v}+\dfrac{1}{20}}$

$\longrightarrow\sf{\dfrac{1}{v}=\dfrac{1}{30}-\dfrac{1}{20}}$

$\longrightarrow\sf{\dfrac{1}{v}=\dfrac{2-3}{60}}$

$\longrightarrow\sf{\dfrac{1}{v}=\dfrac{-1}{60}}$

$\longrightarrow\underline{\underline{\red{\sf{v=-60\;\;cm}}}}$

Calculating size of image

Using formula for magnification produced by Lens

$\longrightarrow\sf{m=\dfrac{v}{u}=\dfrac{h_i}{h_o}}$

$\longrightarrow\sf{\dfrac{v}{u}=\dfrac{h_i}{h_o}}$

$\longrightarrow\sf{\dfrac{(-60)}{(-20)}=\dfrac{h_i}{(10)}}$

$\longrightarrow\sf{3=\dfrac{h_i}{10}}$

$\longrightarrow\underline{\underline{\red{\sf{h_i=30\;\;cm}}}}$

Therefore,

• Position of image will be 60 cm infront of the Lens. ( v = - 60 cm )
• Size of image will be 30 cm, means 3 times enlarged. ( $\sf{\bold{h_o}}$ = 30 cm )
• Nature of Image will be Virtual and erect.

Answer 2

Explanation:

height of object=+10cm

object distance=-20cm *

(object distance is always negative in both the lenses)

focal length=+30cm

*(focal length is always positive in convex lens and negative in concave lens)

Applying lens formula

1/f=1/v-1/u

1/30=1/v-(1/-20)

1/30=1/v+1/20

1/v=1/30-1/20

1/v=-1/60

v=-60 cm

Relation b/w m and v,u

hi/ho=v/u

hi/10=-60/-20

hi=30 cm

Hence, image is enlarged,virtual,erect and forned 60 cm on the left side of the lens