An object is placed 25 cm in front of a lens of focal length 20 cm. 60 cm past the first lens is a second lens of focal length 25 cm. What is the resulting magnification of the object in this setup?

Question

Quizplus · Accepted Answer

First, we can use the thin lens equation for the first lens:
1/f = 1/di + 1/do
where f is the focal length, di is the distance between the object and the lens, and do is the distance between the image and the lens. Plugging in the given values, we get:
1/20 = 1/25 + 1/do
Solving for do, we get:
do = 100 cm

Now we can consider the second lens. Let's use the thin lens equation again, this time using the image from the first lens as the object for the second lens:
1/f = 1/di + 1/do
where f is the focal length, di is the distance between the object and the lens, and do is the distance between the image and the lens. Plugging in the given values, we get:
1/25 = 1/60 + 1/do
Solving for do, we get:
do = 37.5 cm

The overall magnification of the object is the product of the magnification from each lens:
m = (-do/ di) * (-do/ di)
Plugging in the values we found, we get:
m = (-37.5/ 25) * (-100/ 25)
m = 6
Note that the negative signs cancel out because the image is on the opposite side of the lens as the object in both cases. Therefore, the resulting magnification is 6, which corresponds to answer choice A.

An Object Is Placed 25 Cm in Front of a Lens