Question

an object 8cm high is placed 20cm from the converging lens whose focal length is 40cm the magnification in the case is​

Answer 1

Given :

• Height of object, $\sf{h_o}$ = 8 cm
• Position of object, u = -20 cm
• Focal length of converging lens, f = 40 cm

To find :

• Magnification produced, m = ? ; Height of image, $\sf{h_i}$ = ?

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, u is position of object, v is position of image, $\sf{h_i}$ is height of image, $\sf{h_o}$ is height of object and m is magnification produced by lens ]

Solution :

Calculating position of image (v)

Using lens formula

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

$\longrightarrow\sf{\dfrac{1}{40}=\dfrac{1}{v}-\dfrac{1}{(-20)}}$

$\longrightarrow\sf{\dfrac{1}{40}=\dfrac{1}{v}+\dfrac{1}{20}}$

$\longrightarrow\sf{\dfrac{1}{v}=\dfrac{1}{40}-\dfrac{1}{20}}$

$\longrightarrow\sf{\dfrac{1}{v}=\dfrac{1-2}{40}}$

$\longrightarrow\underline{\underline{\red{\sf{v=-40\;\;cm}}}}$

Calculating magnification produced by lens (m)

Using formula for magnification produced by Lens

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

$\longrightarrow\sf{m=\dfrac{(-40)}{(-20)}}$

$\longrightarrow\underline{\underline{\red{\sf{\;\;\;m=2\;\;\;}}}}$

Calculating height of image (hᵢ)

Using formula for magnification produced by Lens

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

$\longrightarrow\sf{2=\dfrac{h_i}{8}}$

$\longrightarrow\underline{\underline{\red{\sf{h_i=16\;\;cm}}}}$

Therefore,

• Magnification produced by the Lens in this case is 2 , and
• Height of image is 16 cm.

Answer 2

Explanation:

Given:

•Height of object, h_{o}=8 cm

•Position of object, u = -20 cm

• Focal length of converging lens, f = 40

cm

To find :

• Magnification produced, m=?

•Height of image, h_{i} = ?

Formulae required :

• Lens formula

${\huge{\underline{\boxed{\mathbb{\blue{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }\purple{}\blue{}\green{}\purple{}}}}}}$

• Formula for magnification produced by Lens

${\huge{\underline{\boxed{\mathbb{\blue{ m = \frac{v}{u} - \frac{ h_{i} }{h_{o}} }\purple{}\blue{}\green{}\purple{}}}}}}$

Solution:

Calculating.position of image (v)

• Using lens formula

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

$= > \frac{1}{40} = \frac{1}{v} - \frac{1}{ - 20}$

$= > \frac{1}{40} = \frac{1}{v} + \frac{1}{20}$

$= > \frac{1}{v} = \frac{1}{40} - \frac{1}{20}$

$= > \frac{1}{v} = \frac{1 - 2}{40}$

$= > v = - 40$

Calculating magnification produced by lens (m)

Using formula for magnification produced by Lens

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

$= > m = \frac{ - 40}{ - 20}$

$= > m = 2$

Calculating height.of image.(h_{i})

Using formula for magnification produced by Lens

$= > m = \frac{hi}{ho}$

$hi = 16$

Therefore,

• Magnification produced by the Lens in this case is 2, and Height of image is 16 cm.