00:01
In this problem, we have been given that there is a concave lens and we have an object here that's placed in front of this concave lens and the height of this object, that's 50 centimeters.
00:15
Further, we are given that the image which is formed as a result of refraction through this concave lens, that's having height of 20 centimeters.
00:27
So, hi represents the height of image and this image is formed on the screen which is placed at a distance of 10 centimeters from the lens.
00:38
So we know that in case of concave lens, the object and the image they are formed on the same side.
00:45
So whatever side we kept the object, the image is formed on the same side of the lens.
00:50
So both are negative in this case because both are lying to the left of this optical center.
00:57
So here the image distance is minus 10 centimeters and we are required to determine the focal length of this lens.
01:08
So to compute that, let's first use these values so that we can relate them and compute the magnification.
01:17
Because magnification is the height of image to the height of object.
01:21
And this is even the image distance with respect to the object distance.
01:25
So from here we can compute the object distance.
01:28
So let's put the values, 20 over 50...