Question

An object of height 2.5 cm is placed at a distance of 10 cm infront of a concave mirror of focal length 5 cm. Find by calculation: (i) the position (ii) the height (iii) the nature of the image formed ​. please give the answer fast and please dont spam

Answer 1

Answer:

Explanation:

being a concave mirrior

f=-5

u=-10

h(height of object)=2.5cm

1/f=1/v+1/u

1/(-5)=1/v+1/(-10)

1/(-5)=1/v-1/10

1/(-5)+1/10=1/v

1-2/10=1/v

-1/10=1/v

-10=v

the image formed is 10 cm in front of the mirrior

the image formed will be real and inverted

height of image/height of object=-v/u

h/2.5=-(-10/-10)

h/2.5=-1

h=-2.5

Answer 2

Answer:

Explanation:

Given,

Object height, h(o) = 2.5 cm

Object distance, u = - 10 cm

Focal length, f = - 5 cm

To Find,

(i) The position

(ii) The height

(iii) The nature of the image formed​.

Formula to be used,

Mirror formula,

1/v + 1/u = 1/f

Magnification formula,

m = - v/u

Solution,

(i) Putting all the values, we get

1/v + 1/u = 1/f

⇒ 1/v + 1/(- 10)= 1/(- 5)

⇒ 1/v - 1/10 = 1/(- 5)

⇒ 1/v = 1/(- 5) + 1/10

⇒ 1/v = - 2 + 1/10

⇒ 1/v = - 1/10

⇒ v = - 10 cm

Hence, the image distance is - 10 cm.

(ii) We know that,

Magnification formula,

m = - v/u

⇒ m = - (- 10)/- 10

⇒ m = - 1

Now,

⇒ m = h(i)/h(o)

⇒ - 1 = h(i)/2.5

⇒ h(i) = 2.5 × (- 1)

⇒ h(i) = - 2.5 cm

Hence, the image height is - 2.5 cm.

(iii) Nature of the image,

The image will be real and inverted.