Question

at what distance from a concave mirror of focal length 10cm , should ab object 3cm high be placed , so that 6cm high erect image is formed

Answer 1

Answer:

Explanation:

Given:-

Focal length ,f = -10cm

Object height ,ho = 3cm

Image height ,hi = 6cm

To Find:-

Object Distance ,u ?

Solution:-

we have to calculate the object Distance . As we know that magnification is ratio of height of image to the height of object.

m = hi/ho

Substitute the value we get

$:\implies$ m = 6/2

$:\implies$ m = 3

Now, we also know that magnification is aslo calculated by the ratio of image distance to the object distance.

m = -v/u

Substitute the value we get

$:\implies$ 3 = -v/u

$:\implies$ v = -3u

Now, using mirror formula1/v + 1/u = 1/f

Substitute the value we get

$:\implies$ 1/(-3u) + 1/u =-1/10

$:\implies$ -1/3u + 1/u = -1/10

$:\implies$ -1 + 3/3u = -1/10

$:\implies$ +2/3u = -1/10

$:\implies$ u = -10×2/3

$:\implies$ u = -20/3

$:\implies$ u = -6.66 cm

Hence, the object distance is 6.66cm from the concave mirror

Answer 2

Answer:

Given :-

• A concave mirror of focal length 10 cm, should an object of 3 cm high be placed, so that 6 cm high erect image is formed.

To Find :-

• What is the object distance.

Formula Used :-

$\clubsuit$ Magnification Formula :

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

where,

• $\sf h_i$ = Height of the image
• $\sf h_o$ = Height of the object

And,

$\longmapsto \sf\boxed{\bold{\pink{m =\: \dfrac{- v}{u}}}}$

where,

• v = Distance of image
• u = Distance of object

$\clubsuit$ Mirror Formula :

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

where,

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

Solution :-

First, we have to find the magnification :

Given :

• Height of the image = 6 cm
• Height of the object = 2 cm

According to the question by using the formula we get,

$\implies \sf m =\: \dfrac{\cancel{6}}{\cancel{2}}$

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

$\implies \sf\bold{\purple{m =\: 3}}$

Again,

Given :

• Magnification (m) = 3

According to the question by using the formula we get,

$\implies \sf 3 =\: \dfrac{- v}{u}$

By doing cross multiplication we get,

$\implies \sf\bold{\purple{v =\: - 3u}}$

Now, we have to find the object distance :

Given :

• Image Distance (v) = - 3u
• Focal length (f) = - 10 cm

According to the question by using the formula we get,

$\dashrightarrow \sf \dfrac{1}{- 3u} + \dfrac{1}{u} =\: \dfrac{1}{- 10}$

$\dashrightarrow \sf \dfrac{- 1 + 3}{3u} =\: \dfrac{1}{- 10}$

$\dashrightarrow \sf \dfrac{2}{3u} =\: \dfrac{1}{- 10}$

By doing cross multiplication we get,

$\dashrightarrow \sf 3u =\: 2(- 10)$

$\dashrightarrow \sf 3u =\: - 20$

$\dashrightarrow \sf u =\: \dfrac{- 20}{3}$

$\dashrightarrow \sf\bold{\red{u =\: - 6.67\: cm}}$

$\therefore$ The object distance is - 6.67 cm .

[ Note : The object should be placed in front of the mirror. ]

$\rule{150}{2}$

#Learn more :

An object is placed 15 cm from a diverging mirror that has a focal length of 10 cm. Its image distance will be?

O 30 cm in front of the mirror

O 30 cm behind the mirror

O 6 cm in front of the mirror

O 6 cm behind the mirror

https://brainly.in/question/39804794