Question

At what distance from a concave mirror of focal length 10 cm should an object 2 cm long be placed in order to get an erect image 6 cm tall ?

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Answer 1

Given that, the focal length of the concave mirror is 10 cm.

{ focal length (f) = -10 cm}

Also given that, the height of the object (ho) is 2 cm and height of the image (hi) = 6 cm.

Using Mirror Formula:

1/f = 1/v + 1/u

1/(-10) = 1/v + 1/u ...........(1st equation)

Also,

m = hi/ho = -v/u

Substitute the known values,

6/2 = -v/u

3 = -v/u

v = -3u ..............(2nd equation)

Substitute value of (2nd equation) in (1st equation)

→ -1/10 = 1/(-3u) + 1/u

→ -1/10 = -1/3u + 1/u

→ -1/10 = (-1 + 3)/3u

→ -1/10 = 2/3u

Cross-multiply them

→ -1(3u) = 2(10)

→ -3u = 20

→ u = -20/3

→ u = -6.67 cm

Now, Substitute the value of u in (2nd equation)

→ v = -3(-6.67)

→ v = +20.01

→ v = 20 cm

{ Image is formed behind the mirror. }

Answer 2

Question -

At what distance from a concave mirror of focal length 10 cm should an object 2 cm long be placed in order to get an erect image 6 cm tall ?

Solution -

In the above Question, the following information is given ...

Focal length = 10 cm

Object Height = 2 cm

Image Height = 6 cm.

Magnification = ( 6 / 2 ) = 3

Now,

Magnification = ( - v / u )

So,

( -v / u ) = 3

=> 3u = -v

=> v = -3u

Also -

According To The mirror Formulae -

$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ = > \dfrac{1}{u} - \dfrac{1}{3u} = \dfrac{1}{10} \\ \\ = > \dfrac{1}{u} (1 - \dfrac{1}{3} ) = \dfrac{1}{10} \\ \\ = > \dfrac{1}{u} = \dfrac{3}{20} \\ \\ = > u \: = \dfrac{20}{3} \approx 6.66 \: cm$

So, the object Distance is ( 20 / 3 ) cm or 6.66 cm

So, the object should be placed at a distance of (20/3) cm from the pole of the mirror to satisfy the above conditions .