00:01
Alright, so the question is an object 10 centimeter high is placed at a distance of 30 centimeters.
00:07
So, i'll write here, so the object height is 10 centimeters.
00:11
So, h1 is 10 centimeters is placed at a distance of 30 centimeters.
00:17
So, u is 30 centimeters from a convex lens of focal length 35.
00:22
So f is equal to 35 centimeters.
00:25
We need to find the position, nature and size of the image.
00:29
Now, as in this case, we know the object distance is taken negative because it is placed on the left hand side.
00:37
So here we have to find the h2 as well as the value of v.
00:43
First we'll calculate the value of v, that is the image distance.
00:48
The lens formula is 1 upon v minus 1 upon u is equal to 1 upon f.
00:54
We need to find 1 upon v.
00:57
So i'll take this minus 1 upon you on the right hand side.
00:59
So 1 upon f plus 1 upon u f is 1 upon 35 u is minus so i'll write here minus 1 upon 30 we'll do the cross multiplication here 30 minus 35 divided by 35 into 30 is 1050 this is 1 upon v 30 minus 35 is minus 5 upon 1050 this is 1 upon v 30 minus 35 is minus 5 upon 1050 is 1 upon v so taking the reciprocal v is equal to 1 .5 .0 .50 divided by minus 5 which is equal to minus 210 centimeters now as is this negative so the image is forming in front of the lens so it is a virtual image because if it was a real image then this negative sign won't be there so this is a virtual image similarly we need to find the h2 so for this the formula is h2 upon h1 is equal to v upon u...