00:01
Hello students, in the first part we need to prove that if the distance of the object do equals to 2f, that is twice the focal length of the lens, and that is twice the focal length of the lens, then the distance of the image, di, will also be equal to 2f.
00:30
Also, the magnification will be minus 1.
00:35
Thus, using the lens formula, 1 upon f equals to 1 upon v minus 1 upon u.
00:46
Here, v is the distance of the image and u is the distance of the object.
00:52
B substitute the values and write it as 1 upon f equals to 1 upon v.
00:58
That is, let's say this is b, i, distance of the image.
01:03
Image minus 1 upon u that is 2 into f however if we take this direction to be the positive direction then it will be minus 2f so we are trying to prove that if u equals to 2f then v is also 2f using the lens formula.
01:30
This can be written as 1 upon distance of the image plus 1 upon 2f.
01:36
This gives us 1 upon distance of the image equals to 1 upon f minus 1 upon 2 f which is equal to 1 upon 2 f.
01:47
Thus we get the distance of the image is also equal to 2 into f.
01:53
Hence proved.
01:56
Also it is given that magnification is minus 1.
02:00
The magnification m is given as minus v upon u, which is minus 2f upon 2f, which is equal to minus 1.
02:13
Now in the second half, what we need to do is we need to perform an experiment.
02:20
We need to the lens and the object in such a way that the image is at a distance of 4f from the object and at a distance of 2f from the lens...