Question

An object of size 2.0 cm is placed at a distance of 20.0 cm in front of a concave mirror of focal length 10.0 cm . find the distance from the mirror at which a screen should be placed in order to obtain sharp image. what will be the nature of image formed ​

Answer 1

Answer:

the height is 10 cm.

Explanation:

the object is virtual and real.

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Answer 2

Answer:

height of object (h)=2 cm

object distance (u)= -20cm

focal length (f)= -10cm

Explanation:

To find :

image distance (v)

height of image (h)

magnification(m)

Formula to be used are

Mirror formula - 1/v + 1/u = 1/f

magnification - m = -v/u =h'/h

solution

putting all values,we get

1/v + 1/u = 1/f

1/v + 1/-20 = 1/-10

1/v = 1/-10 - 1/-20

1v = -20 + 10/200

1v = -1/20

v = -20 cm

hence image distance is -20 cm

Now magnification

m= -v/u = h'/h

h'= h x v/u

h'= 2 x (-20/-20)

h' = -2 cm

m = -v/u

m = 20/-20

m = -1

so the nature is real and inverted .