AP Physics 2: Algebra-Based Flashcards: Images Formed by Lenses
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Review key ideas with interactive flashcards. This set includes 11 cards to help you master important concepts.
For a thin convex lens, on which side of the lens is the focal point where parallel rays converge?
The focal point is on the transmitted side of the lens, meaning the side opposite from where the light rays originated.
Card 1 of 11
All Flashcards (11)
For a thin convex lens, on which side of the lens is the focal point where parallel rays converge?
The focal point is on the transmitted side of the lens, meaning the side opposite from where the light rays originated.
What three variables are related by the thin-lens equation?
The thin-lens equation relates the focal length of the lens ($f$), the distance of the object from the lens ($s_o$), and the distance of the image from the lens ($s_i$).
What is the primary characteristic of a thin concave (diverging) lens regarding parallel light rays?
It refracts incident light rays parallel to the principal axis so they diverge as if they originated from a focal point on the incident side of the lens.
What is the thin-lens equation?
The thin-lens equation is $\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}$, which relates the image distance ($s_i$), object distance ($s_o$), and focal length ($f$).
What is the key difference in how convex and concave lenses refract parallel light rays?
A convex lens converges parallel rays to a real focal point on the transmitted side, while a concave lens diverges parallel rays as if from a virtual focal point on the incident side.
For a thin concave lens, on which side of the lens is the focal point from which parallel rays appear to diverge?
The focal point is on the incident side of the lens, meaning the same side from where the light rays originated.
What is the primary characteristic of a thin convex (converging) lens regarding parallel light rays?
It refracts incident light rays parallel to the principal axis so they converge toward a common focal point on the transmitted side of the lens.
What two factors determine the location of an image formed by a thin lens?
The location of the image depends on the focal length of the lens ($f$) and the distance between the object and the midline of the lens ($s_o$).
An object is placed at a distance $s_o$ from a lens with a known focal length $f$. How would you calculate the location of the image, $s_i$?
You would use the thin-lens equation, $\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}$, and algebraically solve for the image distance, $s_i$.
If you know the object distance ($s_o$) and have calculated the image distance ($s_i$), how can you determine the magnification of the image?
You can calculate the magnification using the formula $M = -\frac{s_i}{s_o}$.
What is the magnification (M) of an image formed by a thin lens?
Magnification is the ratio of the image size ($h_i$) to the object size ($h_o$), given by the equation $M = \frac{h_i}{h_o} = -\frac{s_i}{s_o}$.