Question

. An object of height 2 cm is placed at a distance of
15 cm from a concave mirror of focal length 10 cm
Draw a scale diagram to locate the image. From the
diagram, find the length of the image formed.​

Answer 1

Given :-

Focal length of the concave mirror = 10 cm

Height of the object = 2 cm

Distance of the object = 15 cm

To Find :-

The length of the image formed.

Analysis :-

Here we are given with the focal length, height and the object distance.

Firstly, using the mirror formula substitute the values and find the image distance accordingly.

Then using the magnification formula you can easily find the height of the image.

Solution :-

We know that,

• u = Object distance
• v = Image distance
• m = Magnification
• h = Height
• f = Focal length

Using the formula,

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

Given that,

Focal length (f) = 10 cm

Object distance (u) = 15 cm

Substituting their values,

⇒ 1/v + -1/15 = -1/10

⇒ 1/v = (-3 + 2) / 30

⇒ 1/v = -1 / 30

⇒ v = -30 cm

Using the formula,

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

Given that,

Image distance (v) = -30 cm

Object distance (u) = -15

Object height = 2 cm

Substituting their values,

⇒ Image height / 2 = -(-30)/-15

⇒ Image height / 2 = -2

⇒ Image height = -2 × 2

⇒ Image height = -4 cm

Therefore, the length of the image formed is 4 cm.

Answer 2

Answer:

$➪$ $\sf{\:\large{h_{i}}}$ $=$ $\boxed{\bold{\red{4cm}}}$

Given:

Focal length (f) = 10cm

Object distance (u) = 15cm

Object height ($\sf{h_{o}}$) = 2cm

To find:

Height of image ($\sf{h_{i}}$)

Formula used:

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

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

Solution:

First we have to find image distance (v)

After applying sign convention,

Focal length (f) = -10cm

Object distance (u) = -15cm

($\because$ In front of mirror)

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

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

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

$\sf{⇒\dfrac{1}{v}=\dfrac{10-15}{150}}$

$\sf{⇒\dfrac{1}{v}=-\dfrac{1}{30}}$

$\sf{⇒v = -30 cm}$

Now we have to find height of image by using,

$\sf{➪\:Magnification=\dfrac{h_{i}}{h_{o}}= -\dfrac{v}{u}}$

After applying sign convention,

Object height ($\sf{h_{o}}$) = +2cm

($\because$ Above principal axis)

$\sf{➪\:\dfrac{h_{i}}{h_{o}}= -\dfrac{v}{u}}$

$\sf{➪\:\dfrac{h_{i}}{2}= -\dfrac{\cancel{-}30}{\cancel{-}15}}$

$\sf{➪\:h_{i}= -2 \times 2}$

$\sf{➪\:h_{i}=}$ $\bold{\red{-4cm}}$

∴ Length of image = 4cm (inverted)