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

by Anagha

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by Anagha

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**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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Concave mirror , f is negative , f = -10 cm

u = ?

h = 2 cm

erect image so h' is positive, h' = 6 cm

so m = h'/h = 3

as m = -v/u = 3 => v = -3 u

mirror equation is 1/v + 1/u = 1/f

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

2/(3u) = -1/10

u = -20 /3 cm = -6.67 cm and v = -3 u = 20 cm

v is positive, so image is behind the mirror and is virtual.

in concave mirror , an erect image is obtained when the object is between the pole and the focus. it is magnified and is virtual. so you can verify from values of u, v and f.

u = ?

h = 2 cm

erect image so h' is positive, h' = 6 cm

so m = h'/h = 3

as m = -v/u = 3 => v = -3 u

mirror equation is 1/v + 1/u = 1/f

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

2/(3u) = -1/10

u = -20 /3 cm = -6.67 cm and v = -3 u = 20 cm

v is positive, so image is behind the mirror and is virtual.

in concave mirror , an erect image is obtained when the object is between the pole and the focus. it is magnified and is virtual. so you can verify from values of u, v and f.