pls help with this optics frq for ap physics 2🙏

Question

The figure above represents a glass lens that has one flat surface and one curved surface. After incoming parallel rays pass through the lens, the rays pass through a focal point. The focal length f is the distance from the center of the lens to the focal point.

a. The rays undergo refraction and change direction at the right surface of the lens, as shown. Explain why the angle of refraction of ray 1 is greater than that of ray 2.

b. The index of refraction of the glass is n(glass) and the radius of curvature of the lens's right edge is R. (The radius of curvature is the radius of the sphere of which that edge is a part. A smaller R corresponds to a lens that curves more.) A teacher who wants to test a class's understanding about lenses asks the students if the equation f=n(glass)R makes sense for the focal length of the lens in air. Is the teacher's equation reasonable for determination of the focal length? Qualitatively explain your reasoning, making sure you address the dependence of the focal length on both R and n(glass)

c. An object is placed a distance f/2 (half of the focal length) to the left of the lens. On which side of the lens does the image form, and what is its distance from the lens in terms of f? Justify your answer. (Assume this is a thin lens.)

d. The lens is now placed in water, which has an index of refraction that is greater than air but less than the glass. Indicate below whether the new focal length is greater than, less than, or equal to the focal length f in air.

___ Greater than in air

___ Less than in air

___ The same as in air

Justify your answer qualitatively, with no equations or calculations.

Answer 1

Unfortunately, I'm unable to see the figure, but I can say the following based on what was written:

a) The Law of Refraction is n_A sin θ_A = n_B sin θ_B, where n is the index of refraction and θ is the angle of refraction for that material. Since the product of n and sin θ is a constant, they are inversely proportional. So, if the angle of refraction for A is greater than the index of refraction for B (θ_A > θ_B), that means the index of refraction for A must be less than the index of refraction for B (n_A < n_B). In other words, if the angle goes down, the index must be up

b) A useful way to think about this is imagining light going through a window with no curvature. In that case, the light wouldn't bend going through it. Moreover, a perfectly flat surface corresponds to an infinite radius (meaning you would have to be infinitely far away to see any curvature). However, as soon as the window is bent, the radius becomes smaller and more noticeable, and the light would bend as well. This means that the focal length and the radius are inverse of each other.

As for the the index of refraction, lets go back to the Law of Refraction: n_A sin θ_A = n_B sin θ_B. The index of refraction for a vacuum is 1 and the smallest possible value. Imagine that light is coming from a vacuum and strikes a material that also has an index of 1. The light would not bend in this case, or you could say that it has an infinite focal length (meaning that the light rays converge or diverge infinitely far away). The index for the lens can only increase, and increasing the index deflects the light. The only way it can bend is closer to the lens, meaning that the focal length decreases. Therefore, the focal length length ends up being inversely related to the index of refraction as well!

c) The thin lens equation is 1/s + 1/p = 1/f, where f is the focal length, s is the object distance, and p is the image distance. This equation can be rearranged to solve for the image distance: p = sf / (s-f)The object is placed at a distance of f/2 in front of the lens, which would be our value for s. Inserting that into our image distance equation, we have

p = sf / (s - f) = (f/2 · f) / (f/2 - f) = (f² / 2) / (-f/2) = -f

Alternatively: 1/s + 1/p = 1/f. 1/(f/2) + 1/p = 1/f. 2/f + 1/p = 1/f. 1/f = -1/p, or p = -f

Now as to which side of the lens the image forms, it depends on if we have a convex lens (+f) or a concave lens (-f). A positive image distance is real, meaning the light rays converge, and the image forms on the opposite side of the lens as compared to the original object. A negative image distance is virtual, meaning the light rays diverge and the image forms on the same side of the lens as the original object.

d) Going back to our reasoning in B, we have that the focal length is inversely related to the index of refraction. Going from Air to Glass is an increase in the index of refraction, but going from Water to Glass is a decrease, so as the comparative index of refraction goes down, the focal length goes up! Furthermore, since the light in the glass increases instead of decreasing, the light bends in the opposite way and the focal length ends up being negative