Question

an object 3cm high produces a real image 4.5cm high, when placed at a distance of 20cm from a concave mirror. Calculate: (i) the position of image (ii) focal length of the concave mirror​

Answer 1

Answer:

Object height h1=3cm

image height h2=4.5cm

we know

m=h2/h1=-v/u

so,

4.5/3=-v /-20 (object distance =20)

4.5×-20/3=-v

v=30cm

so image is 30 cm from the mirror.

now , using mirror formula

1/u+1/v=1/f

1/-20+1/-30=1/f

-5/60=1/f

-1/12=1/f

f=-12

Hence , the required focal length is 12cm

Answer 2

Answer:

Height of the object = h = 3 cm

Height of the image = h' = -4.5 cm (Real and Inverted Image)

Object distance from the concave mirror = u = -20 cm

Image distance from the concave mirror = v = ?

Focal length of the concave mirror = f = ?

Firstly, Magnification = h'/h

⇒ m = -4.5/3

⇒ m = -1.5

Now, We also know that, m = -v/u

So, m = -1.5 = -v/(-20)

⇒ -1.5 = -v/-20

⇒ -v = -1.5 × -20

⇒ -v = 30

⇒ v = -30 cm

Now,

Mirror Formula = 1/u + 1/v = 1/f

⇒ 1/-20 + 1/-30 = 1/f

⇒ 1/f = (-30-20)/600

⇒ 1/f = -50/600

⇒ 1/f = -5/60

⇒ 1/f = -1/12

⇒ f = -12 cm

Hence,

1. Image distance from the concave mirror = v = -30 cm
2. Focal length of the concave mirror = f = -12 cm