If you had to straighten out a refracted ray that went through a lens with a focal length of +15 cm, what could you do?
I'm not sure whether to use a lens with a -15 focal length, but is that even possible?
I think it's supposed to be about converging and diverging lenses; does one have a positive focal length and the other a negative one?
Yes, a +15 focal length followed or preceded by a CLOSE -15 focal length lens would result in a "straightened out" beam, and no image would be formed.
To straighten out a refracted ray that went through a lens, you can use the concept of lens equation. The lens equation relates the object distance (denoted as 'u'), the image distance (denoted as 'v'), and the focal length of the lens (denoted as 'f'). It is given by:
1/f = 1/v - 1/u
In this case, the focal length of the lens is +15 cm.
To straighten out the refracted ray, we need to find the image distance ('v') where the ray emerges in a parallel direction to the incident ray. This means that the refracted ray needs to be focused at infinity. Mathematically, this would occur when the image distance ('v') is equal to infinity.
So, we need to find the object distance ('u') that will result in the lens focusing the refracted ray at infinity. Rearranging the lens equation, we have:
1/v = 1/f + 1/u
Since we want 1/v to be zero (to get infinity as the value of 'v'), the equation becomes:
0 = 1/f + 1/u
Substituting the value of the focal length (f = +15 cm) into the equation, we have:
0 = 1/15 + 1/u
To solve for 'u', we can rearrange the equation:
1/u = -1/15
Taking the reciprocal of both sides, we get:
u = -15 cm
Therefore, to straighten out the refracted ray that went through a lens with a focal length of +15 cm, you would position the object at a distance of -15 cm (i.e., 15 cm on the same side as the incident light).