Question

A 10cm tall object is placed perpendicular to the principal axis of convex lenses of focal length 30cm. The distance of the object from the lense is 20cm. Find the: 1) Position, 2) Nature, 3) Size of the image formed. ​

Answer 1

Given

A 10cm tall object is placed perpendicularly to the axis of convex lens of focal length 30cm

To find

1)Position

2)Nature

3)Size of the image formed

Explanation

For a convex lens,

height of the object,ho=10cm

focal length=30cm

object distance=20cm

According to sign conventions,

u=-20 cm

f=30cm

v=?

$\boxed{\bf \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}$

$\dfrac{1}{30} = \dfrac{1}{v} - (\dfrac{ - 1}{20} ) \\ \\ = > \dfrac{1}{30} = \dfrac{1}{v} + \dfrac{1}{20} \\ \\ = > \dfrac{1}{v} = \dfrac{1}{30} - \dfrac{1}{20}$

$= > \dfrac{1}{v} = \dfrac{3 - 2}{60} \\ \\ = > \dfrac{1}{v} = \dfrac{1}{60} \\ \\ = > \boxed{ \sf \: v = 60 \: cm}$

let us calculate the magnification factor

$\boxed{\bf m=\dfrac{v}{u}=\dfrac{h_i}{h_o}}$

$= > \dfrac{60}{-20} = \dfrac{h_i}{10} \\ \\ = > \boxed{\sf h_i =- 30cm}$

Image formed is Real, inverted and enlarged and formed at 30 cm on the other side of lens

Extra information

For concave mirror and biconcave lens

sign conventions are as follows

u=-u

f=-f

v=-v

unknown=+ve

For convex mirror and biconvex lens

sign conventions are as follows

u=-u

f=+f

v=+v

unknown=+ve

Answer 2

AnswEr :

⠀

• Given :

• Focal Length ( f ) = 30cm
• Object Distance ( u ) = - 20cm
• Object Height ( h ) = 10cm

• To Find :

• Position
• Nature
• Size of Image Formed( v )

• Solution :

⠀

Using Mirror Formula :

$\boxed{\red{ \large \boxed {\bf{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }}}}$

• Plugging in the Values.

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

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

$\longrightarrow \bf{ \dfrac{1}{v} = \dfrac{(3 - 2)}{60} }$

$\longrightarrow \bf{ \dfrac{1}{v} = \dfrac{ 1}{60} }$

• The Positive Sign of v shows that the Image is formed at a distance of 60cm to the Right of Optical Centre of the lens.

⠀

_________________________________

$\boxed{\red{\boxed {\bf{Magnification = \frac{v}{ u} }}}}$

$\longrightarrow \bf{M= \dfrac{v}{ -u} }$⠀⠀[A.T.Q.]

$\longrightarrow \bf{M= \dfrac{60}{ -20} }$

$\longrightarrow \bf{M= \cancel\dfrac{-60}{ 20} }$

$\longrightarrow \bf{M = -3}$

⠀

_________________________________

$\boxed{\red{\boxed {\bf{Image\: Height = M \times Object \: Height}}}}$

$\longrightarrow \bf{Image\: Height = - 3 \times 10}$

$\longrightarrow \bf{Image\: Height = -30cm}$

⠀

Negative Sign of Image Height implies that the Image formed is Inverted.

$\large\therefore$ The Image Formed at a Distance of 30cm on the other side of the lens and the Image is Real and Inverted.