Question

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

Answer 1

Answer:

a

The image distance is $v =2.33 cm$

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 $m = 1.553$

e

The image height is $H_i = 2.33cm$

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

            $\frac{1}{f} = \frac{1}{v}+ \frac{1}{u}$

 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

            $v = \frac{f*u}{f-u}$

               $= \frac{4.20 * 1.5}{4.20 - 1.5}$

              $=2.33cm$

Mathematically magnification is represented as

                  $m = \frac{v}{u}$

                     $= \frac{2.33}{1.50} = 1.553$

Mathematically the image height is represented as

                $H_i = H_o * m$

Where $H_i$ is the image height

           $H_o$ is the object height = 1.5

$H_i = 1.5 * 1.553 = 2.33cm$