What Is the Focal Length of the Lens When the Nickel's Image Doubles in Size?

[goWlfpack]

Homework Statement

The nickel's image in the figure below has twice the diameter of the nickel when the lens is 2.90 cm from the nickel. Determine the focal length of the lens.

Homework Equations

M = h'/h = - (p/q)

The Attempt at a Solution

i don't really know how to start this one... eek

 

[Redbelly98]

There is a well-known equation for lenses, try that one.

Hint: it has a "1/f" in it.

 

[thomate1]

I would approach this problem by first understanding the concept of focal length. Focal length is the distance between the center of a lens and its focal point, where parallel light rays converge or diverge after passing through the lens.

In this problem, we are given that the diameter of the nickel's image is twice the diameter of the actual nickel when the lens is 2.90 cm away. This means that the image is magnified by a factor of 2. We can use the magnification equation, M = h'/h = - (p/q), where h' is the height of the image, h is the height of the object, p is the distance between the object and the lens, and q is the distance between the image and the lens.

Since we know that the image is twice the size of the object, we can substitute M = 2 into the equation. We also know that p = 2.90 cm, and we are trying to find the focal length, so q = f. This gives us the equation 2 = - (2.90/f). Solving for f, we get f = -2.90/2 = -1.45 cm.

However, since focal length cannot be negative, we can conclude that the focal length of the lens is 1.45 cm. This means that when an object is placed 1.45 cm away from the lens, its image will be twice the size of the object.