2. An object is 20.0 cm from a converging lens, and the image falls on a screen. When the object is moved 4.00 cm closer to the lens, the screen must be moved 2.70 cm farther away from the lens to register a sharp image. Determine the focal length of the lens.
we have, since the focal length is constant,
1/f = 1/20 + 1/x = 1/16 + 1/(x+2.7)
Just solve for x, and then find f.
To determine the focal length of the lens, we can use the lens formula:
1/f = 1/di - 1/do
Where:
- f is the focal length of the lens
- di is the image distance (distance of the image from the lens)
- do is the object distance (distance of the object from the lens)
Let's assign the following variables:
- di1 is the image distance when the object is 20.0 cm from the lens
- di2 is the image distance when the object is moved 4.00 cm closer to the lens
- do1 is the object distance when the object is 20.0 cm from the lens
- do2 is the object distance when the object is moved 4.00 cm closer to the lens
From the problem statement, we know that:
di1 - do1 = 2.70 cm
di2 - do2 = 0 cm
Since the image falls on a screen, we can assume the screen is located at the image position. Therefore, di1 = 0 cm.
Plugging these values into the lens formula, we get:
1/f = 1/0 - 1/20
1/f = -1/20
To solve for f, we take the reciprocal of both sides:
f = -20 cm
However, the negative sign indicates that the lens is a converging lens. Therefore, the focal length of the lens is +20 cm.