Question

An object of height 3 cm is placed at a distance of 10cm from a convex mirror of focal length 15 cm . Find the position, size and nature of the image.

Answer 1

Given:-

• Object height ,ho = 3cm
• Object distance ,u = -10cm
• focal length ,f = +15cm

To Find:-

• The position, size and nature of the image.

Solution:-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀According to the Question

We have to calculate the image distance & height of image & its nature . Firstly we calculate the image distance .By using mirror formula .

• • 1/v + 1/u = 1/f

where,

• v is the Image Position
• u is the object distance
• f is the focal length

Substitute the value we get

$:\implies$ 1/v + 1/-10 = 1/15

$:\implies$ 1/v -1/10 = 1/15

$:\implies$ 1/v = 1/15+1/10

$:\implies$ 1/v = 2+3/30

$:\implies$ 1/v = 5/30

$:\implies$ 1/v = 1/6

$:\implies$ v = 6cm

• Hence, the image distance is 6 cm.

_________________________________

Now, calculating the height of image by using magnification formula of the mirror.

• m = hi/ho = -v/u

$\longrightarrow$ hi/ho = -v/u

$\longrightarrow$ hi/3 = -6/-10

$\longrightarrow$ hi/3 = 3/5

$\longrightarrow$ hi = 9/5

$\longrightarrow$ hi = 1.8cm

• Hence, the height of image is 1.8 cm & the image formed is virtual & erect .

Answer 2

Answer:

Given :-

• An object of height 3 cm is placed at a distance of 10 cm from a convex mirror of focal length 15 cm.

To Find :-

• What is the position, size and nature of the image.

Formula Used :-

$\clubsuit$ Mirror Formula :

$\longmapsto \sf\boxed{\bold{\pink{\dfrac{1}{f} =\: \dfrac{1}{v} + \dfrac{1}{u}}}}$

where,

• f = Focal Length
• u = Object Distance
• v = Image Distance

$\clubsuit$ Magnification Formula :

$\longmapsto \sf\boxed{\bold{\pink{Magnification\: (m) =\: \dfrac{h_i}{h_o} =\: \dfrac{- v}{u}}}}$

where,

• $\sf h_i$ = Height of the image
• $\sf h_o$ = Height of the object
• v = Image Distance
• u = Object Distance

Solution :-

First, we have to find the image distance or v :

Given :

• Object Distance = - 10 cm
• Focal Length = 15 cm

According to the question by using the formula we get,

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

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

$\dashrightarrow \sf \dfrac{1}{v} =\: \dfrac{1}{15} - \bigg(\dfrac{1}{- 10}\bigg)$

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

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

$\dashrightarrow \sf \dfrac{1}{v} =\: \dfrac{5}{30}$

By doing cross multiplication we get,

$\dashrightarrow \sf 5v =\: 30$

$\dashrightarrow \sf v =\: \dfrac{\cancel{30}}{\cancel{5}}$

$\dashrightarrow \sf\bold{\red{v =\: 6\: cm}}$

$\therefore$ The image distance is 6 cm .

$\rule{150}{2}$

Now, we have to find the height of the image :

Given :

• Height of the object = 3 cm
• Image Distance = 6 cm
• Object Distance = - 10

According to the question by using the formula we get,

$\implies \sf \dfrac{h_i}{3} =\: \dfrac{\cancel{-} 6}{\cancel{-} 10}$

$\implies \sf \dfrac{h_i}{3} =\: \dfrac{\cancel{3}}{\cancel{5}}$

$\implies \sf \dfrac{h_i}{3} =\: \dfrac{3}{5}$

By doing cross multiplication we get,

$\implies \sf 5h_i =\: 9$

$\implies \sf h_i =\: \dfrac{9}{5}$

$\implies \sf\bold{\red{h_i =\: 1.8\: cm}}$

$\therefore$ The height of the image is 1.8 cm and the image is formed is erect and virtual.

$\rule{150}{2}$

#Learn more :

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