# an object is placed at a distance of 10 cm from a convex mirror of focal length 15cm. find the position and nature of the image.

Here,

Object distance, u = -10 cm

Focal length, f = 15 cm

Image distance = v

Now,

1/f = 1/v + 1/u

=> 1/15 = 1/v – 1/10

=> v = 6 cm

So, the image is located at a distance 6 cm from the mirror on the other side of the mirror.

Magnification, m = -v/u = 0.6

The image is virtual, erect and diminished.

• 106

u = -10 cm

f = 15cm

v = ?

m = ?

1/v + 1/ u = 1 / f     (mirror formula )

1/ v +1 / -10 = 1/ 15  (put values and solve it you will get the value of (v) i.e. the position )

1/ v= 1/15 + 1/10

1/v = 2+3/30=5/30=1/6

as 1/ v is 1/ 6 so v = 6 cm

m= -v/u (now put the formula of magnification and solve you will get the value of (m) i.e. nature of the image)

m= -6/-10= 6/10 =3/5

so m = 3/5

• 25

u = -10 cm

f = 15cm

v = ?

m = ?

1/v + 1/ u = 1 / f (mirror formula )

1/ v +1 / -10 = 1/ 15 (put values and solve it you will get the value of (v) i.e. the position )

1/ v= 1/15 + 1/10

1/v = 2+3/30=5/30=1/6

as 1/ v is 1/ 6 so v = 6 cm

m= -v/u (now put the formula of magnification and solve you will get the value of (m) i.e. nature of the image)

m= -6/-10= 6/10 =3/5

so m = 3/5

• 25

can u explain why in this question the focal lenth is taken as POSITIVE ?

• 1

a convex mirror has a virtual focus so thats why it is taken as positive.

• 14
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