Question

2. An object of height 10 cm is kept at a distance of 25 min front of a concave mirror. The
focal length of minor is 15 cm. At what distance rorn the mirror should a screen be kept so
as to get a clear image​

Answer 1

Answer:

Here, object distance= u = -15 cm.

Image distance= v

Focal length=f =-20 cm (concave mirror)

Now, (1/v) + (1/u) = (1/f)

=> (1/v) = (1/f) - (1/u)

=> (1/v)= (-1/20)- (-1/15)

=> (1/v)= (1/60)

=> v = 60 cm

Since (v) is positive , image is formed behind the mirror.

==>Therefore the image is virtual and upright.

Magnification = m = (-v/u) = 4.

Thus, the image has a height 4 times that of the object.

=> Height of image = 4*(height of object)= 4*4 = 16 cm