Question

An object is held 82.7 cm away from a lens. The lens has a focal length of -31 cm.

A.) Determine the image distance

B.) Determine the magnification.

C.) If the object is 4 cm long, determine the image height

D.) Which of the following are true about the image?

The image is virtual

the image is upright

the image is inverted

the image is real

E) Which is true about the lens?

The lens is concave

the lens is converging

The lens is convex

the lens is divering

Answer 1

focal length=f=-31 cm

object distance=u=-82.7 cm

let image distance be v.

then using lens equation:

(1/v)-(1/u)=1/f

==>(1/v)+(1/82.7)=-1/31

==>1/v=(-1/82.7)-(1/31)

==>v=-22.548 cm

part a:

image distance from the lens is 22.548 cm and the image is same side as the object.

part b:

magnification=image distance/object distance=v/u=-22.548/(-82.7)=0.27265

part c:

image height=magnification*object height

=0.27265*4=1.0906 cm

part d:

as image is on the same side of the object, image is virtual.

as height is positive, image is upright.

part e:

as focal length is negative, lens is diverging and convex.