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

heats surface water up

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.