Question

an object of height 100 cm is placed at a distance of 50 cm in front of a convex mirror of focal length is 25 cm find image distance and its height.​

Answer 1

Answer:

Image distance =  16.67 cm behind the mirror

Image height h₂ = 33.33 cm

Explanation:

height of object h₁ =₁ 100 cm

object distance  u= - 50 cm

focal length f = + 25 cm

From Mirror formula

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

$\frac{1}{v}$  - $\frac{1}{50}$ = $\frac{1}{25}$

$\frac{1}{v}$  = $\frac{1}{25}$ + $\frac{1}{50}$  = 3/50

v = 50/3 cm = + 16.67 cm

magnification m = - v/u = h₂/h₁

               Height of image h₂ = -(v/u) x h₁

                                                = - [(50/3)/(-50)] x 100

                                                = + 100/3 cm

                                                = 33.33 cm

Image distance =  16.67 cm behind the mirror

Image height h₂ = 33.33 cm

     

Answer 2

Required Answer :

• The image distance = 16.67 cm
• The height of the image = 33.33 cm

Given :

• Height of object = 100 cm
• Object distance = - 50 cm
• Focal length = 25 cm

To find :

• Image distance
• Height of the image

Solution :

Calculating the image distance :

Using formula :

$\red \bigstar \quad \underline{\boxed{\bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}}$

where,

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

we have,

• u = - 50 cm
• f = 25 cm

Substituting the given values :

$\\ \quad : \implies \sf \dfrac{1}{v} + \dfrac{1}{- 50} = \dfrac{1}{25}$

$\\ \quad : \implies \sf \dfrac{1}{v} - \dfrac{1}{ 50} = \dfrac{1}{25}$

$\\ \quad : \implies \sf \dfrac{1}{v} = \dfrac{1}{25} + \dfrac{1}{50}$

$\\ \quad : \implies \sf \dfrac{1}{v} = \dfrac{2 + 1}{50}$

$\\ \quad : \implies \sf \dfrac{1}{v} = \dfrac{3}{50}$

$\\ \quad : \implies \sf \: v \: = \dfrac{50}{3}$

$\\ \quad : \implies \sf \: v \: = 16.67$

$\red \bigstar \underline{\boxed{\sf{The \: \: image \: \: distance \: = 16.67 \: cm}}}$

Using formula :

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

where,

• m denoted the magnification
• v denotes the image distance
• u denotes the object distance

we have,

• v = 50/3 cm
• u = - 50 cm

Substituting the given values :

$\\ \quad : \implies \sf m = - \dfrac{50}{3} \div - 50$

$\\ \quad : \implies \sf m = - \dfrac{50}{3} \times \dfrac{1}{ - 50}$

$\\ \quad : \implies \sf m = \dfrac{50}{3} \times \dfrac{1}{ 50}$

$\\ \quad : \implies \sf m = \dfrac{1}{3}$

Using formula,

$\red \bigstar \underline{\boxed{\bf m = \dfrac{h}{h'}}}$

where,

• h denotes the object height
• h' denotes the image height

we have,

• m = 1/3
• h = 100 cm

Substituting the given values :

$\\ \quad : \implies \sf \dfrac{1}{3} = - \dfrac{100}{h'}$

$\\ \quad : \implies \sf h' = \dfrac{100}{3}$

$\\ \quad : \implies \sf h' = 33.33$

$\red \bigstar \underline{\boxed{\sf{The \: \: height \: \: of ~~the ~~image \: = 33.33 \: cm}}}$