An object's distance from a converging lens is 3.78 times the focal length. a) Determine the location of the image. Express the answer as a fraction of the focal length.
(b) Find the magnification of the image?
(c) Indicate whether it is upright or inverted?
(d) Indicate whether it is real or virtual?

Respuesta :

Answer:

a) v=1.36f

b)m=.36

c) real

d) Inverted

Explanation:

a)

In this question we have given,

object distance from converging lens,u=-3.78f

focal length of converging lens,=f

we have to find location of image,v=?

and magnification,m=?

we know that u, v and f are related by following formula

[tex]\frac{1}{f} =\frac{1}{v}- \frac{1}{u}[/tex].............(1)

put values of f u in equation (1)

we got,

[tex]\frac{1}{f} =\frac{1}{v}- \frac{1}{-3.78f}[/tex]

[tex]\frac{1}{f}-\frac{1}{3.78f} =\frac{1}{v}[/tex]

[tex]\frac{2.78}{3.78f} =\frac{1}{v}[/tex]

or

v=1.36f

b) Magnification, [tex]m=\frac{v}{u} \\m=\frac{1.36f}{3.78f}\\m=.36[/tex]

c)

Here the object is located further away from the lens than the focal point therefore image is real image and inverted.

d) image is inverted