An object is 15 cm in front of a diverging lens with a
focal length of 10 cm.
a) use ray tracing to determine the location of the image
b) is the image upright or inverted?
c) is the image real or virtual?

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

where are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that , which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

0.19 m/s and 0.45 m/s

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form