00:01
In this question we have a diverging lens which is a concave lens and it looks like this so this is a diverging lens and it is called a diverging lens because when ray of light from an infinite source falls on the lens it gets diverged and it appears to fall on the focus and the focus of our diverging lens is negative because it is opposite to the direction of flow of the light so f for this given diverging lens is equal to minus of 9 .1 centimeters and we have different positions of the object depending upon the position of the object we are supposed to find out the image distance the nature of the image formed and the magnification of the image we know that from lens formula 1 by v minus of 1 by u is equal to 1 by f and u for all cases is negative so we have 1 by v plus 1 by u is equal to 1 by f so in the first instance we have the object distance u to be equal to 27 .3 centimeters so putting this value in the lens formula we get that 1 by v plus 1 divided by 27 .3 is equal to minus of 1 divided by 9 .1 or we can say that 1 by 1 by 9 .1 or we can say that 1 by 1 by v is equal to minus of 1 divided by 9 .1 plus 1 divided by 27 .3 centimeters.
01:49
On making this calculation, we get the value of v to be equal to minus of 6 .825 centimeters.
01:59
Now the negative sign in this distance indicates that the image is formed behind the lens.
02:05
If this is the diverging lens here, the image is formed behind the lens.
02:08
And because it is formed behind the lens so we can say that the image formed is virtual in nature now we're going to find out the value of magnification of the image produced for this case so magnification m is equal to minus of v by u so this becomes equal to 6 .825 whole divided by 27 .3 or this is equal to 0 .2.
02:43
The value of magnification is 0 .25 which means that the image formed is diminished anything less than one we can say that the image formed is diminished and we're gonna proceed the same thing for the second instance where the object distance is equal to 9 .1 centimeters so putting the values we get 1 by v plus 1 divided by 9 .1 is equal to minus of my 9 .1 centimeters or we can say that from this we can say v is equal to minus of 4 .55 centimeters.
03:17
The negative sign again here indicates that the images form behind the lens and it is virtual.
03:27
Now we're going to find out the value of magnification.
03:30
Magnification m would be equal to minus of v by u.
03:33
So it's minus of minus 4 .55 divided by u is equal to 27 is 9 .1 in this case.
03:41
The magnification is always an absolute value and here we get the value of magnification to be equal to 0 .5 which means that the image formed in this case will be exactly half the size of the object in the third case we have the object distance is equal to 4 .55 centimeters so we have 1 by v plus 1 divided by 4 .55 is equal to minus of 1 divided by 9 .1 whole in centimeters or we can say that 1 by is equal to minus of 1 divided by 9 .1 plus 1 divided by 4 .55...