00:01
Hello friends in this question we have a diverging lens diverging lens and diverging lens is a concave lens as it diverges the image and we know that we take the focal length of our concave lens as negative so focal length will be minus of 30 centimeter an object is placed at a distance of 10 centimeters so by default we place the object on the left side of the lens so you will be minus of 10 centimeter that means if we say that here is our focus f, then our object is placed somewhere here.
00:43
This is at a distance u.
00:44
So let me write it down as 4.
00:46
Now let us find out about the nature of the image and the location of the image where the image is formed.
00:52
For this, we are going to use the lens formula, lens formula, which is 1 upon v minus of 1 upon u is equal to 1 upon f, where v is the distance of the image from the lens.
01:11
So if we put the values, we will get 1 upon v.
01:14
Then we have minus of 1 upon you.
01:16
That means it will become plus of 1 upon 10 is equals to minus of 1 upon 30.
01:22
So from here we get 1 upon v equals minus 1 by 30, then minus 1 by 10.
01:28
That is minus of 4 by 30.
01:33
The survey comes out to be minus of 7 .5 centimeters.
01:40
Now let us also find the magnification before drawing the ray diagram let us find the magnification of length is given by v by u so it will be minus of 7 .5 divided by minus of 10 that comes out to be 0 .75 so this is the magnification of the image now if we draw the ray diagram let us draw the ray diagram okay so i'm drawing this figure only let us say this is our object o okay okay let me draw one more figure clearly okay this is our concave lens with the two foci then let us say that here we put the object so one ray will go like this and it will appear to be passing through the focus like this and it will diverge because it is parallel then we have the another ray it will be like this just will we should pass through this point okay this is our ray number two so by drawing these two rays we get an intersection point which is this one so this will be our image i okay so what we did we draw one ray which appears to be passing pass through this focus f1 then we draw a drew one more ray which passes through this point p and they both intersect at this point so here the image will be formed so this is the answer for the a part of this question which was asking to draw the ray diagram so this was the answer for the a part of this question now b part is asking that to find the image distance so we have already found the image distance that is v is equals to minus 7 .5 centimeter so this is the answer for the b part of this question then we'll come to the c part compute the magnification we have also found the magnification which is 0 .75.
03:53
So m is equal to 0 .75 which is the answer for the c part of this question...