In two dimensions, imagine a parabolic mirror like the one shown in red in Figur

Question

In two dimensions, imagine a parabolic mirror like the one shown in red in Figure 20.26, with the equation of the form z = 1/2(1 ? y2). Rays (in green) coming from a semicircle of directions are reflected to become rays (in
blue) parallel to the z-axis (and vice versa).

(a) Show that the yz-vector n = [y, 1]T is normal to the red curve at the point (y, z) = (y, 1/2(1 ? y2)).

(b) If light traveling in the direction [0 ?1]Tstrikes the mirror at (y, z) and reflects, in what direction r does it leave? Write out your answer in terms of y and z.

(c) Show that at (±1, 0) the outgoing ray is in direction [±1 0]T.

(d) Show that incoming rays in direction [0 ?1]Twhose y-coordinate is between ?1 and 1 become outgoing rays in all possible directions in the right half-plane.

(e) If you spin Figure 20.26 about the z-axis, the red curve generates a paraboloid. Show that in this situation, incoming rays in the direction [0 ?1]Twith starting points of the form (x, y, 1), where x2 + y2? 1, produce all possible outgoing rays in the hemisphere of directions with z ? 0.

(f) If we take a second paraboloid defined by spinning z = 1/2(?1 + y2), then arrows in direction [0 1]T, arriving from points of the form (x, y, ?1), where x2 + y2? 1, reflect into outgoing rays in the opposite hemisphere of directions. These two reflections establish a correspondence between (1) two unit disks parallel to the xy-plane, and (2) two hemispheres of directions. Write out the inverse of this correspondence.

(g) Explain how you can represent a texture map on the sphere (e.g., an irradiance map, or an environment map, etc.) by providing textures on two disks.

(h) Estimate the largest value of the change of area for this “dual paraboloid” parameterization of the sphere to show that it uses texture memory quite effectively, even though ?/4 of each texture image (the disk within the unit square) gets used for each half of the parameterization.

Explanation / Answer

2a+11 = 14

Subtract 11 from both sides of the equation

2a + 11 - 11 = 14 - 11

2a + 0 = 3

2a = 3

Divide both sides by 2

2a/2 = 3/2

(2/2)a = 3/2

(1) a = 3/2

a = 3/2

B.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.