0:00
Hello students.
00:02
So in given question the object height, so h1 is given that is a plus 5 centimeter.
00:09
It is away.
00:11
So converging length so we are former this is your converging lens.
00:15
Okay and in converging length the object distance is given that is a 25 centimeter so we are mentioned that is an object and the focal length is given that is a 10 centimeter so we mentioned that is a 10 centimeter that focal length and this is an object distance is given that is a 25 cm.
00:36
So according to a sign convention, so we use a u is consider that is a minus of a 25 centimeter and focal length f is consider that is a 10 centimeter.
00:46
Now there is in that situation we use that is the equations of motion 1 upon v minus 1 upon u is equivalent to a 1 upon f.
00:57
1 upon v that is unknown minus 1 upon u minus 1 upon u is a minus of a 25 is equal to 1 upon f that is a 1 upon 10 centimeter now just change so 1 upon v plus 1 upon 25 is equivalent to a 1 upon 10 just change that equation so 1 upon v 1 upon 10 minus of a 1 upon 25 so lcm is a 50 okay so 10 cancel out of the 5 time and that is a 2 time so dear students after that we get a 1 upon v is equivalent to a 3 by 50.
01:38
So to calculate the position, so v is equivalent to a 50 by 3.
01:43
This is a position of a v.
01:47
Now we calculate the magnification.
01:50
So h1 upon h2 must be equivalent to a v by u...