Assume that light travels from left to right as it passes through the lens. An object 1.50 cm high is placed 1.50 cm in front (to the left) of a lens of focal length 4.20 cm.

(a) Calculate the image distance with THREE SIG FIGS. cm.
(b) The image is
(c) The image is located -
(d) Calculate the magnification, m with THREE SIG FIGS.
(e) Calculate the image height, where the height can be positive or negative depending on the image's orientation (THREE SIG FIGS). cm
(f) The image is

a

The image distance is

b

The image is real and inverted because the image distance in positive

c

The image is located to the left side

d

The magnification is

e

The image height is

f

The image is real and inverted because the image distance in positive

Explanation:

In the Solution We assume that the lens is converging

   The lens Equation is mathematically represented as

 Where f is the focal length = 4.20 cm

            v is the image distance = ?

           u is the object distance = + 1.5 (since is to the left i.e the source of light )

        Making v the subject in the equation above

Mathematically magnification is represented as

Mathematically the image height is represented as

Where is the image height

            is the object height = 1.5

c

explanation:

so easy. just google the electromagnetic waves and it’ll pop up.