Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is theposition of the point where the reflected rays intersect the optic axis?(i) 0.25 m from the mirror.(ii) 0.5 m from the mirror.(iii) 1 m from the mirror(iv) 2 m from the mirror.(v) Infinity, ie. the rays never intersect the optic axis. Problem 2: LensThe lens in the picture (one of the surfaces is flat, light entering from the left) is characterized by tworadiuses of curvature, R1andR2. Which statement about the values ofR1andR2is correct?(i)R1=[infinity],R2<0.(i i)R1=[infinity],R2>0.(iii)R1> 0,R2=[infinity].(iv)R1<0,R2=[inf inity].(v)R1> R2>0.Physics 9B LectureAn alternative derivation is to start with a very wide lens with the two surfaces separated by L, calculate (using the image for the first surface as the object for the second) where an incoming ray parallel to the optic axis eventually crosses the optic axis (a

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

answer and explanation;