Question

an object 4cm in size is placed at a distance of 25cm from a concave mirror of focal length 15 cm find the position nature and height of image

Answer 1

Answer:

Explanation:

Height of the Object(H₀) = 4 cm.

Object Distance(u) = 25 cm.(negative)

Focal Length = 15 cm.(negative)

Using the Mirror's Formula,

   

⇒  

⇒  v = - 37.5 cm.

Thus, the image is formed behind the mirror. Since the Distance is Negative, thus the Image is Real.

Thus, Screen is to be placed at a Distance of 37.5 cm behind the Mirror.

Now, Magnification = H₁/H₀

Also, Magnification = -v/u

  ∴ H₁/H₀ = -v/u

⇒ H₁/4 = - (-37.5)/25

⇒ H₁ = 37.5 × 4/25

⇒ H₁ = 150/25

⇒ H₁ = 6 cm.

Since, the Size of the Image is greater than the size of the Object, thus the image is Magnified.

Answer 2

Answer:

$\displaystyle \red{{ \text{h}_i=6 \ cm}$

Explanation:

Given :

Object distance u = 25 cm

Focal length f = 15 cm

We have mirror formula :

1 / f = 1 / v + 1 / u

- 1 / 15 = 1 / v - 1 / 25

1 / v = 1 / 25 - 1 / 15

5 / v = 1 / 5 - 1 / 3

5 / v = ( 3 - 5 ) / 15

5 / v = - 2 / 15

v = - 75 / 2 = - 37.5 cm

We know :

h_i / h_o = - v / u

Where i and o represent image and object

h_i = - ( - 37.5)  × 4 / 25

h_i = 6 cm

Nature of the image are as :

Real , inverted and magnified.