Question

an object us placed at a distance of 4 cm infront of a convex mirror whose focal length is 5 cm. where is the image formed​

Answer 1

Given :

• distance of object from the pole of convex mirror = 4 cm

• Focal length = 5 cm

To find :

• The position of image

Formula used :

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

the above Formula is mirror formula.

where :-

• u = distance if object from pole of mirror

• v = distance if image from pole of mirror

• f = focal distance

Solution :

To find the position of image we will use mirror formula :-

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

put :-

• u = -4
• f = 5

$\frac{1}{v} + \frac{1}{ (- 4)} = \frac{1}{5}$

$\frac{1}{v} - \frac{1}{ ( 4)} = \frac{1}{5}$

$\frac{1}{v} = \frac{1}{5} + \frac{1}{ 4}$

LCM of 4 and 5 = 20

$\frac{1}{v} = \frac{4 + 5}{20}$

$\frac{1}{v} = \frac{9}{20}$

$v = \frac{20}{9}$

therefore v = 20/9 cm

Image is formed at a distance of 20/9 cm from the pole of mirror ( towards right).

Answer :

Image is formed at a distance of 20/9 cm from the pole of mirror ( towards right).

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Learn more :-

• magnification for mirror = -v/u or height of image/ height of object

• magnification for lens = v/u

• Radius of curvature = 2× focal length

• For convex mirror focal length is always positive.

• For concave mirror focal length is always negative.

lens Formula :-

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

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Answer 2

$\huge\sf { \dag Question }$

An object is placed at a distance of 4 cm in front of a convex mirror whose focal length is 5 cm. Where is the image formed?

$\huge\sf { \dag Solution }$

${\underline {\underline {\rm {\red { Given }}}}}$

• $\sf\orange { u ( object \: distance )= - 4cm }$

• $\sf\orange { f ( focal \: length )= 5cm }$

${\underline {\underline {\rm {\red { To \: find }}}}}$

• $\sf { v ( position \: of \: the \: image )}$

${\underline {\underline {\rm {\red { Calculation }}}}}$

We know that,

${\huge {\boxed {\tt { \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}}}$

Where,

• v is the image distance
• u is the object distance
• f is focal length

Substituting values to find the position of the Image-

$\sf\red { \qquad \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }$

$\sf { \qquad \leadsto \dfrac{1}{v} + \dfrac{1}{ (-4)} = \dfrac{1}{5} }$

$\sf { \qquad \leadsto \dfrac{1}{v} + \dfrac{1}{(-4)} = \dfrac{1}{5} }$

$\sf { \qquad \leadsto \dfrac{1}{v} = \dfrac{1}{5} - \dfrac{1}{(-4) } }$

$\sf { \qquad \leadsto \dfrac{1}{v} = \dfrac{1}{5} + \dfrac{1}{4 } }$

$\sf { \qquad \leadsto \dfrac{1}{v} = \dfrac{9}{ 20} }$

$\sf { \qquad \leadsto v= \dfrac{20}{9} cm}$

[ After reciprocating both sides ]

Therefore, the image formed at 20/9 cm in the right from the pole.

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$\huge\sf { \dag Additional \: Information }$

Mirror Formula:

• ${\boxed {\tt { \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}}$

Where,

|| f = R/2 ||

and

F = focal length or distance between pole.

v = image distance from the pole

u = object is from the pole

R = distance between centre of curvature and pole.

Magnification :

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

Where,

$\tt { h_i}$ = image height from principal axis

$\tt { h_o}$ = object height from principal axis

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