00:05
Hello, we are given a height of the image as no more than 2 centimeters, which is 0 .0 2 meters.
00:15
The height of object is 2 meters.
00:21
So, the identification is the height of image, the height of object.
00:29
2 .2 over 2, which is 0 .0 .0.
00:33
There is a modification.
00:37
But we also define notification as image distance over object distance.
00:44
So image distance because multiplication times object distance.
00:48
Image distance is 0 .0 of object distance.
00:56
So u is an object distance and v is an image distance.
01:13
So now our focalite is giving us 5 centimeters which is 0 .05 meters and so we use the range from the one over u plus one over b because one over a half...