Question

Heya!
Please answer the following physics numericals:
1. An object is placed at distance of 10cm from a convex mirror of focal length 15cm. Find the position and nature of the image. Draw the ray diagram.
2. An object 7cm is placed at 27cm in front of a concave mirror of focal length 18cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and nature of the image. Draw the ray diagram.
3. A convex mirror used for rear view on an automobile has a radius of curvature of 3.00m. If a bus is located at 5.00m from the mirror. Find the position, nature and
size of the image.
Please answer these questions with explanation.

Answer 1

Number (1) :-

• u = -10 cm

• f = 15 cm

Using mirror formula :-

1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

⇒ 1/v = 1/15 - (-1)/10

⇒ 1/v = 5/30

⇒ v = 6 cm

Postion of image : 6 cm from mirror.

Nature : virtual, erect, diminished.

Number (2) :-

• hₒ = 7 cm

• u = -27 cm

• f = -18 cm

Again using mirror formula :-

⇒ 1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

⇒ 1/v = (-1)/18 - (-1)/27

⇒ 1/v = (-3 + 2)/54

⇒ 1/v = -1/54

⇒ v = -54 cm

Now, according to the formula of Magnification of a spherical mirror :-

m = -(v/u) = hᵢ/hₒ

⇒ -(-54/-27) = hᵢ/7

⇒ -2 = hᵢ/7

⇒ hᵢ = -2(7)

⇒ hᵢ = -14 cm

Required distance = 54 cm

Size of image : 14 cm

Nature of image : Real, enlarged, inverted

Number (3) :-

• R = 3 m

• u = -5 m

Focal length of the mirror :-

= R/2

= 3/2 m

According to mirror formula :-

⇒ 1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

⇒ 1/v = 1/(3/2) - (-1)/5

⇒ 1/v = 2/3 + 1/5

⇒ 1/v = 13/15

⇒ v = 15/13

⇒ v = 1.15 m

From the formula of magnification :-

m = -(v/u)

⇒ m = -(1.15/-5)

⇒ m = + 0.23

Position of image : 1.15 m from the mirror.

Size : Diminished by a factor of 0.23 .

Nature : virtual, erect .