Answer:
1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))
Explanation:
Problem 1
For this problem we use two equations the equations of the focal distance in mirrors
f = r / 2
f = 2/2
f = 1 m
The builder's equation
1 / f = 1 / o + 1 / i
Where f is the focal length, "o and i" are the distance to the object and the image respectively.
For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0
1 / f = 0 + 1 / i
i = f
i = 1 m
The image is formed at the focal point
The correct answer is iii
Problem 2
For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat
Case 1 Flat lens - convex (convergent)
R₂ = infinity
R₁ > 0
Cas2 Flat-concave (divergent) lens
R₂ = infinity
R₁ <0
Why the correct answers are iii and iv
Problem 3
For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface
1 / f₂ = 1 / (L-f₁) + 1 / i
1 / i = 1 / f₂ - 1 / (L-f₁)
1 / i = (L-f₁-f₂) / f₂ (L-f₁)
i = f₂ (L-f₁) / (L - (f₁ + f₂))
This is the image of the rays that enter parallel to the first surface
