Question

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Answer 1

Question :

An object placed of height of 5 cm placed 40 cm in front of a concave lens of focal length 20 cm. Find the position and size of the image formed.

Answer :

• Position of the image or image distance is (-13.4) cm
• Height of the image is 3.35 cm.

Explanation :

Given :

• Height of the object, ho = 5 cm
• Object distance, u = 40 cm
• Focal length of the mirror, f = 20 cm

To find :

• Position of the image = ?
• Height of the image, ho = ?

Knowledge required :

Lens formula :

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

Where,

• f = Focal length of the mirror
• v = Image distance
• u = Object distance

Formula for magnification :

$\boxed{\sf{m = \dfrac{v}{u} = \dfrac{h_{I}}{h_{o}}}}$

Where,

• v = Image distance
• u = Object distance
• hi = Image height
• ho = object height
• m = Magnification

[Note : By sign convention, in case of a concave mirror, the focal length and the Object Distance are taken as negative]

Solution :

First let us find the image distance :

By using the lens formula and substituting the values in it, we get :

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

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

$:\implies \sf{(-)\dfrac{1}{20} = \dfrac{1}{v} + \dfrac{1}{40}}$

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

$:\implies \sf{\dfrac{(-2) - 1}{40} = \dfrac{1}{v}}$

$:\implies \sf{\dfrac{(-3)}{40} = \dfrac{1}{v}}$

$:\implies \sf{v = \dfrac{40}{(-3)}}$

$:\implies \sf{v = -13.4(approx.)}$

$\boxed{\therefore \sf{v = (-13.4)\:cm}}$

Hence the image distance is (-13.4) cm.

Now let's find the height of the image :

By using the formula for Magnification and substituting the values in it, we get :

$:\implies \sf{m = \dfrac{v}{u} = \dfrac{h_{I}}{h_{o}}}$

$:\implies \sf{\dfrac{v}{u} = \dfrac{h_{I}}{h_{o}}}$

$:\implies \sf{\dfrac{(-13.4)}{(-20)} = \dfrac{h_{I}}{5}}$

$:\implies \sf{\dfrac{13.4}{20} = \dfrac{h_{I}}{5}}$

$:\implies \sf{13.4 \times 5 = 20h_{I}}$

$:\implies \sf{67 = 20h_{I}}$

$:\implies \sf{\dfrac{67}{20} = h_{I}}$

$:\implies \sf{3.35 = h_{I}}$

$\boxed{\therefore \sf{h_{i} = 3.35\:cm}}$

Hence the image height is 3.35 cm.

Answer 2

Answer:

Question :

An object placed of height of 5 cm placed 40 cm in front of a concave lens of focal length 20 cm. Find the position and size of the image formed.

Answer :

Position of the image or image distance is (-13.4) cm

Height of the image is 3.35 cm.

Explanation :

Given :

Height of the object, ho = 5 cm

Object distance, u = 40 cm

Focal length of the mirror, f = 20 cm

To find :

Position of the image = ?

Height of the image, ho = ?

Knowledge required :

Lens formula :

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

f

1

=

v

1

−

u

1

Where,

f = Focal length of the mirror

v = Image distance

u = Object distance

Formula for magnification :

\boxed{\sf{m = \dfrac{v}{u} = \dfrac{h_{I}}{h_{o}}}}

m=

u

v

=

h

o

h

I

Where,

v = Image distance

u = Object distance

hi = Image height

ho = object height

m = Magnification

[Note : By sign convention, in case of a concave mirror, the focal length and the Object Distance are taken as negative]

Solution :

First let us find the image distance :

By using the lens formula and substituting the values in it, we get :

\begin{gathered}:\implies \sf{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}} \\ \\\end{gathered}

:⟹

f

1

=

v

1

−

u

1

\begin{gathered}:\implies \sf{\dfrac{1}{(-20)} = \dfrac{1}{v} - \dfrac{1}{(-40)}} \\ \\\end{gathered}

:⟹

(−20)

1

=

v

1

−

(−40)

1

\begin{gathered}:\implies \sf{(-)\dfrac{1}{20} = \dfrac{1}{v} + \dfrac{1}{40}} \\ \\\end{gathered}

:⟹(−)

20

1

=

v

1

+

40

1

\begin{gathered}:\implies \sf{(-)\dfrac{1}{20} - \dfrac{1}{40} = \dfrac{1}{v}} \\ \\\end{gathered}

:⟹(−)

20

1

−

40

1

=

v

1

\begin{gathered}:\implies \sf{\dfrac{(-2) - 1}{40} = \dfrac{1}{v}} \\ \\\end{gathered}

:⟹

40

(−2)−1

=

v

1

\begin{gathered}:\implies \sf{\dfrac{(-3)}{40} = \dfrac{1}{v}} \\ \\\end{gathered}

:⟹

40

(−3)

=

v

1

\begin{gathered}:\implies \sf{v = \dfrac{40}{(-3)}} \\ \\\end{gathered}

:⟹v=

(−3)

40

\begin{gathered}:\implies \sf{v = -13.4(approx.)} \\ \\\end{gathered}

:⟹v=−13.4(approx.)

\begin{gathered}\boxed{\therefore \sf{v = (-13.4)\:cm}} \\ \\\end{gathered}

∴v=(−13.4)cm

Hence the image distance is (-13.4) cm.

Now let's find the height of the image :

By using the formula for Magnification and substituting the values in it, we get :

\begin{gathered}:\implies \sf{m = \dfrac{v}{u} = \dfrac{h_{I}}{h_{o}}} \\ \\ \end{gathered}

:⟹m=

u

v

=

h

o

h

I

\begin{gathered}:\implies \sf{\dfrac{v}{u} = \dfrac{h_{I}}{h_{o}}} \\ \\ \end{gathered}

:⟹

u

v

=

h

o

h

I

\begin{gathered}:\implies \sf{\dfrac{(-13.4)}{(-20)} = \dfrac{h_{I}}{5}} \\ \\ \end{gathered}

:⟹

(−20)

(−13.4)

=

5

h

I

\begin{gathered}:\implies \sf{\dfrac{13.4}{20} = \dfrac{h_{I}}{5}} \\ \\ \end{gathered}

:⟹

20

13.4

=

5

h

I

\begin{gathered}:\implies \sf{13.4 \times 5 = 20h_{I}} \\ \\ \end{gathered}

:⟹13.4×5=20h

I

\begin{gathered}:\implies \sf{67 = 20h_{I}} \\ \\ \end{gathered}

:⟹67=20h

I

\begin{gathered}:\implies \sf{\dfrac{67}{20} = h_{I}} \\ \\ \end{gathered}

:⟹

20

67

=h

I

\begin{gathered}:\implies \sf{3.35 = h_{I}} \\ \\ \end{gathered}

:⟹3.35=h

I

\begin{gathered}\boxed{\therefore \sf{h_{i} = 3.35\:cm}} \\ \\\end{gathered}

∴h

i

=3.35cm

Hence the image height is 3.35 cm.