Question

An object 5cms in sides is kept at 15 cms from concave mirror of focal length 20 cms at what distance from mirror should a screen be placed so that tobget a sharp image also finde nature and size of image​

Answer 1

Given : The distance of object (h¹) is 5cm & Distance from the object (u) is -15cm & Concave mirror of focal length (f) is -20cm.

To Find : Find the size or height of the mirror (h²) ?

__________________________

Solution : By applying formula we can calculate the size of image as given in the question that. An object of 5cm in size in placed at a distance of 15cm from it concave mirror of focal length of 20cm.

$~$

$\underline{\frak{As~ we ~know~ that~:}}$

• $\boxed{\sf\pink{\dfrac{1}{v}~+~\dfrac{1}{u}~=~\dfrac{1}{f}}}$★

$~$

$\pmb{\sf{\underline{According ~to ~the ~Given~ Question~:}}}$

$~$

$\qquad{\sf:\implies{\dfrac{1}{v}~+~\dfrac{1}{u}~=~\dfrac{1}{f}}}$

$\qquad{\sf:\implies{\dfrac{1}{v}~+~\dfrac{1}{(- - 15)}~=~\dfrac{1}{- 20}}}$

$\qquad{\sf:\implies{\dfrac{1}{v}~=~\dfrac{1}{15}~-~\dfrac{1}{20}}}$

$\qquad{\sf:\implies{\dfrac{1}{v}~=~\cancel\dfrac{300}{5}}}$

$\qquad:\implies{\underline{\boxed{\frak{\purple{\pmb{V~=~60~cm}}}}}}$★

$~$

Therefore,

• The screen should be placed 60 cm in front of the mirror.

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◗Now, calculate the size of the mirror :

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$\qquad{\sf:\implies{\dfrac{h^1}{h^2}~=~(\dfrac{v}{u})}}$

$\qquad{\sf:\implies{\dfrac{h^2}{5}~=~\dfrac{(60)}{(- 20)}}}$

$\qquad:\implies{\underline{\boxed{\frak{\pink{\pmb{h^2~=~3~cm}}}}}}$★

$~$

Hence,

$\therefore\underline{\sf{The~height~of~the~image (h^2)~is~\bf{\underline{\pmb{3~cm}}}~\sf{\&}~\sf{the~screen~placed~at~the~mirror (v)~is~\bf{\underline{\pmb{60~cm}}}}}}$

Answer 2

Given : The distance of object (h¹) is 5cm & Distance from the object (u) is -15cm & Concave mirror of focal length (f) is -20cm.

To Find : Find the size or height of the mirror (h²) ?

__________________________

Solution : By applying formula we can calculate the size of image as given in the question that. An object of 5cm in size in placed at a distance of 15cm from it concave mirror of focal length of 20cm.

~

\underline{\frak{As~ we ~know~ that~:}}

As we know that :

\boxed{\sf\pink{\dfrac{1}{v}~+~\dfrac{1}{u}~=~\dfrac{1}{f}}}

v

1

+

u

1

=

f

1

★

~

\pmb{\sf{\underline{According ~to ~the ~Given~ Question~:}}}

According to the Given Question :

According to the Given Question :

~

\qquad{\sf:\implies{\dfrac{1}{v}~+~\dfrac{1}{u}~=~\dfrac{1}{f}}}:⟹

v

1

+

u

1

=

f

1

\qquad{\sf:\implies{\dfrac{1}{v}~+~\dfrac{1}{(- - 15)}~=~\dfrac{1}{- 20}}}:⟹

v

1

+

(−−15)

1

=

−20

1

\qquad{\sf:\implies{\dfrac{1}{v}~=~\dfrac{1}{15}~-~\dfrac{1}{20}}}:⟹

v

1

=

15

1

−

20

1

\qquad{\sf:\implies{\dfrac{1}{v}~=~\cancel\dfrac{300}{5}}}:⟹

v

1

=

5

300

\qquad:\implies{\underline{\boxed{\frak{\purple{\pmb{V~=~60~cm}}}}}}:⟹

V = 60 cm

V = 60 cm

★

~

Therefore,

The screen should be placed 60 cm in front of the mirror.

~

◗Now, calculate the size of the mirror :

~

\qquad{\sf:\implies{\dfrac{h^1}{h^2}~=~(\dfrac{v}{u})}}:⟹

h

2

h

1

= (

u

v

)

\qquad{\sf:\implies{\dfrac{h^2}{5}~=~\dfrac{(60)}{(- 20)}}}:⟹

5

h

2

=

(−20)

(60)

\qquad:\implies{\underline{\boxed{\frak{\pink{\pmb{h^2~=~3~cm}}}}}}:⟹

h

2

= 3 cm

h

2

= 3 cm

★

~

Hence,

\therefore\underline{\sf{The~height~of~the~image (h^2)~is~\bf{\underline{\pmb{3~cm}}}~\sf{\&}~\sf{the~screen~placed~at~the~mirror (v)~is~\bf{\underline{\pmb{60~cm}}}}}}∴

The height of the image(h

2

) is

3 cm

3 cm

& the screen placed at the mirror(v) is

60 cm

60 cm