Question

The front surface of a glass cube 4.60 cm on each side is placed a distance of 35.0 cm in front of a converging mirror that has a focal length of 19.0 cm.

(3 PARTS)

A. Where is the image of the front surface of the cube located? [ Answer in CENTIMETERS]

B. What is the image characteristics?

     Check all that apply

     Real,   Virtual,   Upright,    Inverted,    Magnified,   Reduced

C. Is the image of the cube still a cube?

    Yes or No?

    

Answer 1

part a:

for the image of the front surface:

using cartesian sign convention for the converging mirror:

object distance=u=-35 cm

focal length=f=-19 cm

let image distance be v.

then using mirror equation:

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

==>(1/v)-(1/35)=-1/19

==>v=-41.562 cm

so image is 41.562 cm from the mirror, on the same side as the object.

part b:

image is real.

as real image by a converging mirror,, it is inverted.

magnification=-v/u=-1.1875

hence it is magnified

part c:

as object placed at different distances for a converging mirror gets magnified by different amount, the image wont be a cube.