Midterm With Answer Key - Image Computation | CS 510, Exams of Computer Science

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CS510 Midterm, Spring 1999

Name: _____ ANSWER KEY ________

ID # _______________________

Question Pts Max

Total 100

1. True/False Questions. Circle the appropriate answer and elaborate if you feel necessary. (10 Points)

(a) true/false: A Lambertian Surface reflects equal amounts of light in all directions. Thus it is sometimes called a perfect diffuse scatterer. This also means that a Lambertian Surface appears equally bright from all viewpoints.

Think about it, the surfaces looks darker as it turns away from the viewer.

(b) true/false: The Uniform Adaptive Sampling ray tracing algorithm shoots more than one ray into the scene for each pixel. Moreover, it shoots the same number of rays into the scene for each pixel.

Adaptive means it adapts the number of rays to the detected changes about the pixel.

(c) true/false: Precisely determining the infinite set of 3D points that are within a solid defined using Constructive Solid Geometry (CSG) model can be complex. Consequently, one “goal oriented” approach is to use ray-tracing to actually determine which points are on the visible surfaces of CSG models.

This is true as was explained in lecture and the text.

(d) true/false: The aperture on a camera controls the amount of light reaching the lens.

The primary reason for having an aperture, particularly a varying aperture, is to control the amount of light.

(e) true/false: The extrinsic parameters of a camera describe properties not just of the camera, but of the camera and its relationship to a scene.

For example, the pose of the camera with respect to the scene is an extrinsic parameter.

3. Phong Illumination. (15 Points)

In this course we have developed the following equation for the illumination of a point on a surface.

The last term represents the cosine of the angle between vector R and V raised to the ns power. The difficulty with using this formula in practice is that the vector R is not immediately known. Instead, it is necessary to derive a general expression for R in terms of vectors that are known: i.e. N and L ,

You may refer to the following figure for clarification about the definitions N, L and R.

Please provide the expression for vector R in terms of N and L.

R = L + 2 D = L + 2(( L N N⋅ ) − L)

To understand this construction, first note that the scalar L dot N is a number between zero and one representing the projection of the point L onto the unit vector N. If N is scaled by this amount, then the result is a point along N at the same vertical height as L. Call this point A, and observe that A=(L•N)N. Next, let the vector D run from the tip of L to this point A. The reflection vector R that we want may be described by going first to the tip of L and then in the direction (and length) of D twice. Since D runs from L to A, D=A-L= (L•N)N-L. Put this all together and you get the equation above.

I k Sa d I (^) a f (^) i di I (^) l i k Sd d N L k Ss s R V n i

n s

λ =^ λ +^ (^ )^ λ [^ λ(^ ⋅ ) +^ λ(^ ⋅ ) ]

=

1

D D

4. Color Spaces. (15 Points)

a) Explain in simple terms why CRT monitors emit light in only three narrow bands centered about the following approximate wavelengths: 450 nm, 550 nm and 650 nm.

These frequencies match the peak receptivity of the blue, green and red color receptors for the human retina.

b) Define each of the components of HSV color space relative to RGB color space. While you need not use an equation in every case, be as precise as possible.

H = Hue is angle about axis [1,1,1] in RGB cube. Red is the zero degree point.

S = Saturation is proportional to the distance to edge of cube. S equals zero is gray

and S equals 1 is at the edge of the cube: a saturated color.

V = Value is projective distance up the [1,1,1] axis. This is commonly called

intensity. The formula is:

V

R B G

A B⋅ in place ofa bx x + a by y +a bz z

6. Ray Intersection. (30 Points)

Compute the 3D point I where the ray originating at point S and passing through point T intersects the plane defined by points A, B, and C. To solve this problem, you will first derived the general solution, and then provide the solution for a specific case:

a) For the general solution, let:

Provide a simple linear algebraic expression for the intersection point. To accomplish this, you will want to introduce some intermediate terms (following the approach set forth in class). Do not expand forms unless necessary, i.e. you can write:

The point of intersection is the position along the ray at t star, where t star is the t value for the intersection of the ray with the plane. Thus,

I = R t( ) = S + t (T −S)

To define t star, first we need equation for the plane. The first step is to find the unit normal to the plane. This is defined by two vectors known to lie in the plane

N^ √ B^ A^ C^ A

B A C A

( − ) × ( − )

( − ) ×^ ( − ) Next, determin the offset D of the plane from the origin.

N ⋅ A =D

The plane equation is therefore

N

x

y

z

⋅ =D

The intersection of the plane is found by substituting the ray equation into the plane equation and solving for t star N ⋅ R t( ) = D, ⇒ N ⋅ (^) ( S + t T( −S )) = D ⇒ N S⋅ + t N( ⋅ (T −S )) =D Therefore:

t

D N S

N T S

S

s

s

s

T

t

t

t

A

a

a

a

B

b

b

b

C

c

c

c

x

y

z

x

y

z

x

y

z

x

y

z

x

y

z

= = = = =

b) What is the actual point of intersection for the following case:

S = T = A = B = C=

0

0

10

1

1

0

1

1

1

2

2

1

2

0

1

Follow the previousgeneral derivation :

By Inspection

Therefore

Finally

I t

B A C A

N N A D D

t

I

= +

−

( − ) =^ ( − ) = −

= ⋅ = ⇒ =

=

−

−

=

= +

−

=

0

0

10

1

1

10

1

1

0

1 1 0 0 0 1

1

1 10

10

9

10

0

0

10

9

10

1

1

10

9 10 9 10 1

,

√ (^) ,.