00:01
So for part a, we're going to use the lens formula, 1 over do, plus 1 over d -i is equal to 1 over f.
00:12
Now, f is equal to given as 0 .3, but because this is, this will be a diverging lens, therefore, we will say f is equal to negative 0 .30 meters.
00:27
And the image formed is a virtual image.
00:30
Therefore, d .i will be equal to negative 0 .24 meters.
00:36
So now we can start putting some numbers in.
00:39
So we're looking for 1 over do plus 1 over negative 0 .24 is equal to 1 over negative 0 .3.
00:51
So 1 over do would be equal to negative 1 over 0 .3 and then minus or 1 over 1.
01:01
Negative 0 .24 and the negative and the negative make it positive so solving this for do gives do is equal to 1 .2 meters and now for part b so for part b we know that magnification is equal to the absolute value of d .i over d o and that is equal to so, d .i was equal to 0 .24 and d .o was equal to 1 .2.
01:36
So that is equal to 0 .2...