Question

An object 9cm height is placed at a distance of 3cm from a convex mirror of focal length 6cm.Find the 1) Position 2) Nature 3) Sized of the image formed​
plz I need the answer

Answer 1

★ Provided Question ★

An object 9cm height is placed at a distance of 3cm from a convex mirror of focal length 6cm.Find the

• 1) Position
• 2) Nature
• 3) Sized of the image formed.

★ Required Solution ★

1.Position of the image:

${\underline {\underline {\rm { Given: } }}}$

• Object distance (u) = –3cm (left of the mirror)
• Focal length (f) = +6cm (as it is convex mirror)

${\underline {\underline {\rm { To \: calculate:} }}}$

• Image distance (v) i.e Position of the image

${\underline {\underline {\rm { Calculation: } }}}$

We'll apply here mirror formula to find the position of the image. We know that:

• ${\boxed {\huge {\bf {\pink { \dfrac{1}{v}+ \dfrac{1}{u} = \dfrac{1}{f} }}}}}$

Substituting values:

$\rm { :\implies \dfrac{1}{v}+ \dfrac{1}{-3} = \dfrac{1}{+6} }$

$\rm { :\implies \dfrac{1}{v}- \dfrac{1}{3} = \dfrac{1}{6} }$

$\rm { :\implies \dfrac{1}{v}= \dfrac{1}{6} +\dfrac{1}{3} }$

$\rm { :\implies \dfrac{1}{v}= \dfrac{1+2}{6} }$

$\rm { :\implies \dfrac{1}{v}= \dfrac{3}{6} }$

$\rm { :\implies v= \dfrac{6}{3} }$

$\rm \red { :\implies v=+ 2 cm } </p><p></p><p>Therefore, the position of image is <strong>2 cm</strong> behind the convex mirror. </p><h2>_____________________________________</h2><p></p><h3>ii) Nature of the image</h3><p></p><p>Since the image is formed behind the convex mirror its nature will be virtual and erect. </p><p></p><p></p><h2>___________________________________</h2><p></p><h3>iii) Size of the image</h3><p></p><p>To find the size of the image we need to calculate the magnification first. </p><p></p><p>So, We know that:</p><p></p><ul><li>[tex]{\boxed {\huge {\bf {\pink { m= -\dfrac{v}{u} }}}}}$

Where,

• m = magnification =?
• u = object distance = –3cm
• v = image distance = +2cm

Now, substituting values:

$\rm { :\implies m= -\dfrac{(+2)}{(-3)} }$

$\rm { :\implies m= \dfrac{2}{3} }$

$\rm { :\implies m= 0.6}$ $\orange { \bigstar }$

With the another formula of magnification,

• ${\boxed {\huge {\bf {\pink { m= \dfrac{h_i}{h_o} }}}}}$

Where,

• m = magnification = +0.6
• h_i = height of the image=?
• h_o = height of the object= 9cm

Substituting values-

$\rm { :\implies +0.6= \dfrac{h_i}{9} }$

$\rm { :\implies h_i = 9 \times 0.6}$

$\rm \red { :\implies h_i = 5.4cm}$

Therefore, size of the image is 5.4 cm.

___________________________________

Answer 2

Given :-

Height of the object = 9 cm

Distance of the object = 3 cm

Focal length of the convex mirror = 6 cm

To Find :-

Position of the image behind the convex mirror.

Nature of the image.

Sized of the image formed.

Analysis :-

Firstly, using the mirror formula you can get the position of the image accordingly.

Next find the position of the image accordingly.

Finally, using the formula of magnification you can find the size of the image formed easily.

Solution :-

We know that,

• f = Focal length
• u = Distance between mirror and object
• m = Magnification
• v = Distance between mirror and image

Using the formula,

$\underline{\boxed{\sf Mirror \ formula=\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}$

Given that,

Distance between mirror and object (u) = -3 cm

Distance between mirror and image (v) = 6 cm

Substituting their values,

1/v = 1/-3 + 1/6

1/6 = 1/v - 1/3

1/v = 1/6 + `1/3

1/v = 1+2/6

1/v = 3/6

1/v = 2 cm

Therefore, the the position of image behind the convex mirror is 2 cm.

According to the question,

The image is formed behind the convex mirror so the nature will be virtual and erect.

Using the formula,

$\underline{\boxed{\sf Magnification =-\dfrac{Image \ distance}{Object \ distance} }}$

We know that,

• v = Image distance
• u = Object distance

Given that,

Image distance (v) = 2 cm

Object distance (u) = -3 cm

Substituting their values,

m = - 2/-3

m = 2/3

m = 0.6 cm

Now, we have to

$\underline{\boxed{\sf Magnification = \frac{h_i}{h_o} }}$

We know that,

• hᵢ = Object height
• hₒ = Image height

Given that,

Magnification (m) = 0.6 cm

Image height (hₒ) = 9 cm

Substituting their values,

0.6 = hᵢ/9

hᵢ = 1.6 × 9

hᵢ = 4.5 cm

Therefore, the size of the image formed is 4.5 cm.