A diverging lens of focal length 18.0m is used to view a shark that is 90.0m away from the lens. If the image formed is 1.0m long, calculate the: (i) image distance; (ii) length of the shark.

i. + 22.5 m ii. 4.0 m

Explanation:

i. Image distance

Using the lens formula

1/u + 1/v = 1/f where f = focal length = + 18.0 m, u = object distance = distance of shark away from lens = + 90.0 m and v = image distance from lens = unknown

So, we find v

1/v = 1/f - 1/u

= 1/+18 - 1/+90

= (5 - 1)/90

= 4/90

v = 90/4

= + 22.5 m

So the image is real and formed 22.5 m away on the other side of the lens.

ii Length of Shark

Using the magnification formula, m = image height/object height = image distance/object distance. image height = 1.0 m where object height = length of shark.

m = image distance/object distance

= v/u

= +22.5/+90

= 0.25

0.25 = image height/object height

So,

object height = image height/0.25

= 1.0 m/0.25

= 4.0 m

So, the length of the shark is 4.0 m

answer: strength i guess

explanation: