00:01
Given that we an object and the height of the object is 4 centimeter.
00:07
This is the height of the object and it is placed 7 centimeter from the to the left from a converging lens.
00:15
So here we have a converging lens and its focal length is given as 12 centimeter and from this converging lens we are placing the object which is placed at 7 centimeter away from this and towards the left.
00:30
The height of the object is given as h4 which is 4 centimeter and we are also placing a diverging lens.
00:38
We are placing a diverging lens of focal length minus 14 centimeter which is 2 centimeter away towards the right of the converging lens.
00:49
Now we have to find the image, we have to find the position and the height of the final image.
00:58
So from this diagram we can say that we know from this information we can say that since u is less than focal length, u is less than the distance of focal length then the image will be virtual.
01:14
We can say that the image will be virtual, the image is virtual for converging lens.
01:23
So from the lens formula 1 minus v 1 by v minus 1 by u is equal to 1 by f.
01:29
Now focal length of the converging lens is positive so this will be 1 by v plus 1 by 7 is equal to 1 by 12 and from here we can calculate v and it comes out to be minus 12 into 7 by 5 centimeter.
01:47
So we have calculated and v is negative, v is negative that simply suggests that the image is virtual.
01:53
So we can also find the magnification for this image let us say that magnification is m this will be given by v by u and it is minus 12 into 7 by 5 multiplied by the yeah this is the distance and multiplied by the image multiplied by the image distance that we had and which is 7 centimeter.
02:16
So if we cancel out we will get it will comes out to be 12 by 5 and we can say this way this is simply 2 .4.
02:28
Now if we have calculated magnification then we know that magnification is also equal to height of image by height of object.
02:35
So from here we will put 2 .4 and the height of image will comes out to be 4 into 2 .4 where 4 is the height of object that we had this is height of image.
02:48
So height of image it will be 9 .6 centimeter.
02:53
Now this is the image for the this is the height of the image that is formed due to the converging lens this is imaginary because height is positive.
03:03
Now this imaginary object sorry this we can say that this virtual not imaginary this virtual this virtual image this virtual image will work as object for diverging lens we can say for diverging lens for the second lens diverging lens diverging lens.
03:38
So the diagram we can see from the diagram that we had converging lens here and image object we have put it here...