00:01
So we're looking at an object that's placed in front of a diverging lens, and we want to find out the characteristics of the image.
00:08
So we're looking for the position in size and whether it's real, virtual, upright, or inverted.
00:15
So what we can start with is we're given the magnitude of the focal length is 20 centimeters, and since we have a diverging lens, the focal length that we're going to use is negative in the calculations.
00:30
So we're going to use this when we first go ahead and find the image position, and we'll do that by using the thin lens equation, which states that 1 over p or the object distance from the lens, plus 1 over q, the image distance that we're looking for is equal to 1 over f.
00:53
So we're told that p is 40 centimeters, so we can go ahead and solve for q.
00:59
And again we're using f equals to negative 20 centimeters.
01:03
And we'll find that q or the image distance is negative 13 .3 centimeters.
01:11
So because we have a negative value, this tells us that it's in front of the lens and it's going to be a virtual image.
01:22
So we know where the image will be and we know that it's virtual.
01:27
So now what we want to find is the size of the image and whether it's upright, or inverted.
01:34
So what we're going to do is we're going to go ahead and use the magnification equation, which is negative q over p...