At what distance from a concave mirror of focal length 10cm should an object 2 cm long replace an erect image?

Given that focal length, f= -10cm

Magnification, m= Image height/ Object height= 6/2 = 3

Also magnification= - v/u

3= -v/u

v= -3u

As 1/u + 1/v = 1/f

1/u + 1/-3u = 1/ -10

Therefore u= -6.667

So object should be placed at a distance of 6.667 cm in front of the mirror.

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At what distance from a concave mirror of focal length 20 cm should an object 2 cm long be placed in order to get an erect image 8 cm tall?

Height of the object 'ho' = 2 cm

Focal length of the mirror 'f' = −10 cm
Height of the image 'hi' = 6 cm
We have to find the distance of the object from the mirror 'u'.
Using the magnification formula, we get 

`m= h_i/h_o=-v/u` 

`m=6/2=-v/u` 

`m=3=-v/u` 

thus v=-3u 

Now, using the mirror formula, we get

 `1/f=1/v+1/u` 

`1/-10=1/(-3u)+1/u` 

`1/10=1/(3u)-3/(3u) =(-2)/(3u)` 

`u=(-2xx10)/3` 

u=-6.67 cm  

Thus, the distance of the object from the mirror 'u' is −6.67 cm.

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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