Question

3.
An object of size 7em is placed
27cm in front of concave mirror of focal
length 18cm At what distance should
the screen be placed so that a sharp
focussed image can be obtained?Find
the nature of image
14cm Imageis real inverted & same
side of object
-14cm,image is real, inverted and
same side of object
14cm,image is imaginary errect and
on opposite side of object
14cm,image is virtual errected and
on opposite side of object​

Answer 1

Given:-

• Object -height ,ho = 7cm

• Object distance ,u = -27cm

• Focal length, f = -18cm

To Find:-

• Image distance ,v

Solution:-

Using mirror formula

• 1/v + 1/u = 1/f

Substitute the value we get

→ 1/v + 1/(-27) = -1/18

→ 1/v - 1/27 = -1/18

→ 1/v = -1/18 +1/27

→ 1/v = -3 + 2/54

→ 1/v = -1/54

→ v = -54 cm

Here, negative sign show the image formed is real and Inverted.

So ,the object should be placed 54 cm From the object to get real image.

Now,

•magnification , m = hi/ho = -v/u

Substitute the value we get

→ hi/ho = -v/u

→ hi/7 = -54/(-27)

→ hi/7 = 54/27

→ hi = 7×2

→ hi = 14 cm

Therefore, the height of the image is 14 cm.