An image of a candle is produced by a convex lens 14 cm away. If the focal length of the lens is 7 cm, how far from the lens is the candle placed?
(USe the thin lens equation)
For this question, how would you know whether the image is virtual or real?(in order to know if the distance to the mirror is negative or positive)
Solve
1/do + 1/di = 1/do + 1/14 = 1/f = 1/7
The answer is that the object distance is (do) = 14. You get a 1:1 real image.
The image is real because do is positive.
To determine the distance from the lens to the object, you can use the thin lens equation:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance from the lens (positive for a real image, negative for a virtual image)
u = object distance from the lens (positive for an object on the same side as the incident light, negative for an object on the opposite side)
Given:
f = 7 cm
v = 14 cm
We can rearrange the thin lens equation to solve for u:
1/u = 1/f - 1/v
Substituting the given values, we get:
1/u = 1/7 - 1/14 = (2 - 1)/14 = 1/14
Now, we can find u by taking the reciprocal of both sides:
u = 14 cm
Therefore, the candle is placed 14 cm away from the lens.
To determine whether the image formed by the convex lens is virtual or real, you need to consider the sign convention. In this case, if the object distance (u) is positive, the image distance (v) will also be positive for a real image. If the object distance (u) is negative, the image distance (v) will be negative for a virtual image. In our calculation, since the object distance (u) is positive, we can conclude that the image formed is real.