Question

an object 5cm is placed at a distance of 10cm in front of convex mirror of radius of curvature 30 cm find the position nature and size the image​

Answer 1

Given :

• Height of the object = 5 cm

• Object Distance = 10 cm

• Radius of curvature = 30 cm

To find :

• Position of the image/ or Image Distance (v)

• Nature of the Image

• Size of the image/ or Height of the Image (hi)

Solution :

First let us find the focal length of the mirror :

Using the relation ;

$\boxed{\bf{f = \dfrac{R}{2}}}$

Where :

• f = Focal length
• R = Radius of curvature

Using the above relation by substituting the values of it , we get :

$:\implies \bf{f = \dfrac{R}{2}}$

$:\implies \bf{f = \dfrac{30}{2}}$

$:\implies \bf{f = 15}$

$\boxed{\therefore \bf{focal\:length\:(f) = 15\:cm}}$

Hence, the focal length of the convex mirror is 15 vm

To find the Image distance (v) :

We know the mirror formula i.e,

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

Where :

• v = Image distance
• u = Object Distance
• f = Focal length

Using the mirror formula and substituting the values in it, we get :

$:\implies \bf{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}$

Here,

• u = (-10) cm
• f = 15 cm

[Note :- Here, the object distance of the mirror is taken as negative because the object distance in a convex mirror is always taken as negative]

$:\implies \bf{\dfrac{1}{15} = \dfrac{1}{v} + \dfrac{1}{(-10)}}$

$:\implies \bf{\dfrac{1}{15} - \dfrac{1}{(-10)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{1}{15} + \dfrac{1}{10} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{2 + 3}{30} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{5}{30} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{1}{6} = \dfrac{1}{v}}$

$:\implies \bf{6 = v}$

$\boxed{\therefore \bf{Image\:Distance\:(v) = 6\:cm}}$

Hence , the image distance or the position of the mirror is 6 cm.

Height of the image :

First let us find the Magnification of the mirror :

Using the formula for magnification and substituting the values in it, we get :

$\boxed{\bf{m = - \dfrac{v}{u}}}$

Where :

• m = Magnification
• v = Image Distance
• u = Object Distance

Now, using the formula for magnification and substituting the values in it, we get :

$:\implies\bf{m = - \dfrac{(-6)}{10}}$

$:\implies\bf{m = \dfrac{6}{10}}$

$:\implies\bf{m = 0.6}$

$\boxed{\therefore \bf{Magnification\:(m) = 0.6}}$

Hence, the magnification of the mirror is 0.6.

Now , using the other formula for magnification i.e,

$\boxed{\bf{m = \dfrac{h_{i}}{h_{o}}}}$

Where :

• m = Magnification
• hi = Height of image
• ho = Height of object

$:\implies \bf{0.6 = \dfrac{h_{i}}{5}}$

$:\implies \bf{0.6 \times 5 = h_{i}}$

$:\implies \bf{3 = h_{i}}$

$\boxed{\therefore \bf{Height\:of\:image\:(h_{i}) = 3\:cm}}$

Hence, the Image height is 3 cm.

Nature of the image :

The image formed is virtual , upright and smaller than the object.

Answer 2

$\green{\LARGE\underline{\underline{\sf Given:}}}$

• Object Height = 5 cm
• Object Distance = 10 cm
• Radius of curvature = 30 cm

$\green{\LARGE\underline{\underline{\sf Find:}}}$

• Position of the image
• Nature of the Image
• Size of the image

$\green{\LARGE\underline{\underline{\sf Solution:}}}$

Here,

✫ u = -10cm

✫ v = ?

✫ $\bold{h_i}$ = ?

✫ R = +30cm

we, know that

$\boxed{\sf f = \dfrac{R}{2} }$

where,

• R = +30cm

So,

➢ $\sf f = \dfrac{R}{2}$

➢ $\sf f = \cancel{\dfrac{30}{2}}$

➢ $\sf f = +15cm$

$\sf \therefore f = + 15cm$

____________________

we, know that mirror formula is given by

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

where,

• u = -10cm
• f = +15cm

So,

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

$\sf \to \dfrac{1}{v} + \dfrac{1}{ - 10} = \dfrac{1}{ + 15}$

$\sf \to \dfrac{1}{v} - \dfrac{1}{ 10} = \dfrac{1}{15}$

$\sf \to \dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{ 10}$

$\sf \to \dfrac{1}{v} = \dfrac{2 + 3}{30}$

$\sf \to \dfrac{1}{v} = \dfrac{5}{30}$

$\sf \to 5v = 30$

$\sf \to v = \cancel{\dfrac{30}{5}} = + 6cm$

The image is formed at a distance of 6cm.

______________________

Now, magnification of (m) mirror is given by

$\boxed{\sf m = \dfrac{ - v}{u} }$

where,

• v = 6cm
• u = -10cm

So,

$\longrightarrow\sf m = \dfrac{ - v}{u}$

$\longrightarrow\sf m = \cancel{\dfrac{ -6}{ - 10}} = \dfrac{3}{5}$

$\longrightarrow\sf m = \dfrac{3}{5} = 0.6$

Also,

$\boxed{ \sf m = \dfrac{ h_{i}}{ h_{o}} }$

where,

• m = 0.6
• $\bold{h_o}$ = 5cm

So,

$\implies\sf m = \dfrac{ h_{i}}{ h_{o}}$

$\implies\sf 0.6= \dfrac{ h_{i}}{5}$

$\implies\sf 0.6 \times 5=h_{i}$

$\implies\sf h_{i} = 3.0cm$

$\underline{\sf \therefore height \: of \: the \: image \: is \: 3cm}$

______________________

Thus, the image is formed behind the convex mirror and its nature is virtual and erect.