Question

An object of 4 cm high is placed at a distance of 6 cm. In front of a concave mirror of focal length 12cm.Find the position ,nature and size of the image.

Answer 1

Answer:

P and F , we get virtual and erect

m=-v/u

=-12/-6

=+2

also m=hi/h0

2=hi/4cm

hi=8cm

hence image is magnified , that is 8cm height.

Answer 2

Position of image is 12 cm, nature of image is virtual and erect and size of image is 8 cm.

Solution

We have,

• Height of object(h_o) = 4 cm.
• Position of object(u) = -6 cm[As it's in front of image]
• Focal length (f) = -12 cm.

We will use mirror formula to find position of image(v) : -

◕ 1/f = 1/v + 1/u

✒ 1/-12 = 1/v + 1/-6

✒ -1/12 = 1/v - 1/6

✒ 1/v = 1/6 - 1/12

✒ 1/v = (2 - 1)/12

✒ 1/v = 1/12

✒ v = 12 cm

∴ Position of image = 12 cm

Now using magnification formula :

◕ m = -v/u

✒ m = -12/-6

✒ m = 2 cm.

∴ Nature of image is virtual and erect.

Now we will use magnification formula :

◕ m = -v/u = h_i/h_o

✒ -12/-6 = h_i/4

✒ 2 = h_i/4

✒ h_i = 4 × 2

✒ h_i = 8 cm

∴ Size of image = 8 cm.