Question

An object of height 2cm is placed at 20cm distance in front of a concave mirror whose focal length is 15cm . calculate the height

Answer 1

Explanation:

Given:-

• Height of the object $\rm (h_{o})$ = 20cm.

• Object distance (u) = 20cm

• Focal length (f) = 15cm

• Type of mirror used - Concave

To Find:-

• The height of the image $\rm (h_{i})$

Solution:-

For a concave mirror:-

• Object distance (u) = -ve

• Focal length (f) = -ve

By using mirror formula:-

$\boxed{ \sf \dashrightarrow \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }$

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

$\sf \longrightarrow \dfrac{1}{v} = \dfrac{1}{ - 15} - \dfrac{1}{ - 20}$

$\sf \longrightarrow \dfrac{1}{v} = \dfrac{1}{ - 15} + \dfrac{1}{ 20}$

$\sf \longrightarrow \dfrac{1}{v} = \dfrac{ - 4 + 3}{ 60}$

$\sf \longrightarrow \dfrac{1}{v} = \dfrac{ - 1}{ 60}$

$\sf \longrightarrow v= -60cm$

$\sf \therefore \underline{Image \: distance \: (v) = -60cm}$

$\sf \boxed{\sf\dashrightarrow Magnification = \dfrac{-v}{u} = \dfrac{h_{i}}{h_{o}}}$

$\rm \longrightarrow \dfrac{-v}{u} = \dfrac{h_{i}}{h_{o}}$

$\rm \longrightarrow \dfrac{-( - 60)}{ - 20} = \dfrac{h_{i}}{2}$

$\rm \longrightarrow - 3 = \dfrac{h_{i}}{2}$

$\rm \longrightarrow h_{i} = - 6$

Note:-

Here negative sign indicates that the image is inverted.

Therefore:-

Height of the image = -6cm

Answer 2

Given :-

☄ Height of the object, h = 20cm.

☄ Object distance, u = 20cm

☄ Focal length, f = 15cm

To Find :-

The height of the image

Solution :-

using mirror formula

$\bigstar$ A mirror formula may be defined as the formula which gives the relationship between the distance of image (v), distance of object (u), and the focal length (f) of the mirror. It may be written as:

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

Where,

v = Distance of image from optical centre of lens

u = Distance of object from optical centre of lens

and f = Focal length of lens

For a concave mirror

Object distance, u = -ve

Focal length, f = -ve

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

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

${\sf{ \leadsto \dfrac{1}{v} = \dfrac{1}{ - 15} - \dfrac{1}{ - 20}}}$

${\sf {\leadsto\dfrac{1}{v} = \dfrac{1}{ - 15} + \dfrac{1}{ 20}}}$

${\sf {\leadsto \dfrac{1}{v} = \dfrac{ - 4 + 3}{ 60}}}$

${\sf {\leadsto \dfrac{1}{v} = \dfrac{ - 1}{ 60}}}$

$\sf \leadsto v= -60cm$

thus, image distance, v = -60cm.

we know,

${\sf {\boxed{\bf Magnification = \dfrac{-v}{u} = \dfrac{h'}{h}}}}$

${ \leadsto{\sf \dfrac{-v}{u} = \dfrac{h'}{h}}}$

${\sf {\leadsto\dfrac{-( - 60)}{ - 20} = \dfrac{h'}{2}}}$

${\sf { \leadsto - 3 = \dfrac{h'}{2}}}$

${\bf {\leadsto h' = - 6cm}}$

thus, height of image is -6cm

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