Question

1. Using the thin lens equation (1/f=(1/s)+(1/s')), prove that the focal length of a lens is equal to the image distance when there is parallel incident light

2. There is a specific case where for a single converging lens the object distance equals the image distance. Find the relation between the focal length and the object distance in this case.

Answer 1

1) for parallel incident rays the object considered at infinity

S = infinity

1/S + 1/S' = 1/f

1/infinity + 1/S' = 1/f

0 + 1/S' = 1/f

f = S' = image distance

2) S = S'

1/S + 1/S = 1/f

2/S = 1/f

f = S/2

Answer 2

Part 1 : Let s : denotes image distance

s' : denotes object distance

f: focal length

Now 1/f = (1/s) +(1/s')

Since there is a parallel incident => s ' = infinity => 1/s' = 0

=> 1/f = (1/s) +(1/s')

=> 1/f = (1/s)+0

=> 1/f = 1/s => f = s

=> Focal length = image distance

Part 2:

1/f = (1/s) +(1/s')

Here we have image distance = object distance => s = s'

=> 1/f = (1/s) +(1/s') = > 1/f = (1/s') + (1/s') => 1/f = 2/s'

=> s' = 2f

Thus Object distance = 2* focal length

Answer 3

Part 1)

When incoming light is parallel, the object distance is considered to be infinity.

The formula to apply is the thin lens equation

1/f = 1/p + 1/q

p is the object distance, and 1 divided by infinity is zero, thus 1/p cancels.

We are left with 1/f = 1/q

Invert and thus f = q

Focal Length equals image distance

Part 2)

Apply the thin lens equation again

1/f = 1/p + 1/q

We want p to equal q, so substitute

1/f = 1/p + 1/p

1/f = 2/p

Cross multiply

p = 2f

f = .5p

Thus the focal lenght is half the object distance