A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50cm tall and inverted.

Note: This is the complete question; A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Explanation:

The required formulas are;

1. lens formula given by, 1/s' + 1/s = 1/f

2. magnification = h'/h = s'/s

where h' is image height, h is object height, s' is image distance, s is object distance, f is focal length of lens.

h' = 4.50cm, h = 3.20cm, f = 70cm ( f of a converging lens is positive)

solving for s' and s in eqn (2)

4.50/3.20=s'/s

1.4=s'/s

s'=1.4s

to find the values of s' and s, we use equation (1) and substitute s' = 1.4s

1/1.4s + 1/s = 1/70

2.4/1.4s =1/70

s = 2.4*70/1.4 = 120cm

s' = 1.4*120 = 160cm

Therefore, the object is on the left, 120cm from the lens while the image is 160cm on the right of the lens

Since the object is at a distance between f and 2f from the lens, the image formed is real, inverted and magnified

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-29T16:38:36+02:00

We have that the object is 120cm to the LHS of the lens, and the image Resides 160cm to the RHS of the lens

s= 120cm

s'= 160cm

What is are the positions of the object and the image?

Question Parameter(s):

A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object

The image is 4.50cm tall and inverted.

Generally, the equation for the Magnification is mathematically given as

h'/h = s'/s

Therefore

4.50/3.20=s'/s

s'=1.4s

Generally, the equation for the lens is mathematically given as

1/s' + 1/s = 1/f

Therefore

1/1.4s + 1/s = 1/70

2.4/1.4s =1/70

s = 2.4*70/1.4

s= 120cm

Also

s' = 1.4*120

s'= 160cm