Question

An object of height 6 cm is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. Find position, nature and height of an image.Solve the given problem.
( Ans : v =-30 cm, height = 12 cm, Real, inverted and magnified

Answer 1

Focal length(f) = 10 cm.

Object distance(u) = 15 cm.

Height of the Object = 6 cm.

Using the Mirror's formula,

1/f = 1/v + 1/u

⇒ 1/-10 = 1/v + 1/-15

⇒ 1/v = 1/-10 + 1/15

∴ 1/v = (-3 + 2)/30

⇒ v = -30 cm.

Since, image distance is negative, therefore, it is real and inverted.

Now, magnification = -v/u

= -(-30)/15

= 2

Since, magnification in greater than 1, therefore Image is magnified.

Also, m = H of image/H of object.

⇒ 2 = H of image/6

∴ Height of image = 12 cm.

Characteristics of Image are ⇒

Real, Inverted and Magnified.

Hope it helps.

Answer 2

Answer:

Position of image: Beyond Centre of Curvature (C)

Nature of image: Real and Inverted

Size: Enlarged

And height of image= 12cm

Explanation:

Given

Height of object (h)= 6cm

Object distance (u)= -15cm

Focal length (f)= -10cm

W.K.T 1/f=1/v +1/u

Therefore, 1/-10= 1/v-1/15

We will get, v=-30cm

Again we have

Linear magnification (m)=h¹/h= -v/u

h¹/6= -(-30)/-15

h¹/6= -2

h¹= -12cm

Therefore

Height of the image (h¹) = 12cm