00:01
Part a of the given problem, the position of the final image and the magnification is given by writing the lenses formula that is 1 over p plus 1 over q is equal to 1 over f, where magnification is minus q over p.
00:24
So then our total magnification will be, total magnification will be m1 magnification 1 times magnification 2.
00:33
So we'll use these formulas and get our total magnification in the position of the image.
00:40
So we'll first find a q.
00:43
So q we can find by substituting value for p, that is 1 over 40, plus 1 over q is equal to 1 over 30.
00:50
From here our cue will be 120 centimeters and our modification 1 will be minus 120 divided by 40.
01:00
This gives us minus 3.
01:04
And then our new position of the object will be minus 120 minus 110.
01:14
This gives us minus 10 centimeter.
01:17
F is minus 20.
01:19
Then our new q will be substituting those values in m lenses equation or new q will be 20 centimeter and then modification m2 will be minus 20 divided by minus 10 this gives us magnification of a 2 then our total modification will be m1 times in 2 as we said earlier so when to find those two numbers minus 3 times 2 this gives us so the final image is positioned at a 20 centimeter to the right of the second diverging lens.
02:05
And the magnification is of course minus six...