A lens of focal length 15cm forms on erect image which is three times the size of the object. The distance between the object and the image is ___.
Answer Details
We can use the lens formula, 1/f = 1/v - 1/u, where f is the focal length of the lens, v is the distance between the lens and the image, and u is the distance between the lens and the object.
From the problem, we know that the focal length of the lens is 15 cm, and the image is erect and three times the size of the object. This means that the image distance v is positive and the object distance u is negative (since the object is in front of the lens).
Let's assume that the object distance u is -x cm, where x is a positive number. Then, the image distance v is +3x cm, since the image is three times the size of the object. Substituting these values into the lens formula, we get:
1/15 = 1/(+3x) - 1/(-x)
Simplifying the right-hand side, we get:
1/15 = (1 + 3)/3x
Multiplying both sides by 3x, we get:
3x/15 = 4
Simplifying, we get:
x = 20
Therefore, the distance between the object and the lens is -20 cm (since it is in front of the lens), and the distance between the image and the lens is +60 cm (since it is behind the lens). The distance between the object and the image is the sum of these distances, which is:
(-20) + (+60) = 40 cm
Therefore, the answer is 40cm.