## Computer-Graphics

Question 1 |

Which of the following statements is/are True regarding the solution to the visibility problem in 3D graphics ?

S1 : The Painter’s algorithm sorts polygons by depth and then paints (scan - converts) each Polygon onto the screen starting with the most nearest polygon. S2 : Backface Culling refers to eliminating geometry with back facing normals.**Code :**

S1 only | |

S2 only
| |

Both S1 and S2 | |

Neither S1 Nor S2 |

Question 1 Explanation:

__Visibility problem in 3D graphics__

1. Painter's algorithm

- A depth sorting method

- Surfaces are sorted in the order of decreasing depth

- Surfaces are drawn in the sorted order, and overwrite the pixels in the frame buffer - Subtle difference from depth buffer approach: entire face drawn

- Two problems:

• It can be nontrivial to sort the surfaces

• There can be no solution for the sorting order

2. Back Face Culling

- Back faces: faces of opaque object which are “pointing away” from viewer

- Back face culling – remove back faces (supported by OpenGL)

How to detect back faces

- If we find backface, do not draw, save rendering resources

- There must be other forward face(s) closer to eye

- F is face of object we want to test if backface

- P is a point on F

- Form view vector, V as (eye – P)

- N is normal to face F

3. View-Frustum Culling

- Remove objects that are outside the viewing frustum

- Done by 3D clipping algorithm (e.g. Liang-Barsky)

4. Ray Tracing

- Ray tracing is another example of image space method

- Ray tracing: Cast a ray from eye through each pixel to the world

5. Z(Depth buffer algorithm)

Reference: https://web.cs.wpi.edu/~emmanuel/courses/cs543/slides/lecture08_p2.pdf

Question 2 |

Memory mapped displays

are utilized for high resolution graphics such as maps | |

uses ordinary memory to store the display data in character form | |

stores the display data as individual bits | |

are associated with electromechanical teleprinters |

Question 2 Explanation:

● Graphs can be displayed on a screen by writing character values into a special area of RAM within the video controller.
● Prior to cheap RAM that enabled bit-mapped displays, this character cell method was a popular technique for computer video displays.

Question 3 |

While making bubbled lists, which of the following options are available?

Square,disc,tringle | |

Triangle,disc,circle | |

Triangle,square,circle | |

Disc,square,circle |

Question 3 Explanation:

→ Disc: A filled circle

→ Circle : An unfilled circle

→ square : A filled square

→ Circle : An unfilled circle

→ square : A filled square

Question 4 |

MPEG involves both spatial compression and temporal compression. The spatial compression is similar to JPEG and temporal compression removes_____frames.

Voice | |

Spatial | |

Temporal | |

Redundant |

Question 4 Explanation:

→ Joint Photographic Experts Group (JPEG) is a method to compress pictures and graphics.

→ The JPEG process involves blocking, the discrete cosine transform, quantization, and lossless compression.

→ Moving Pictures Experts Group (MPEG) is a method to compress video.

→ MPEG involves both spatial compression and temporal compression. The former is similar to JPEG, and the latter removes redundant frames.

→ The JPEG process involves blocking, the discrete cosine transform, quantization, and lossless compression.

→ Moving Pictures Experts Group (MPEG) is a method to compress video.

→ MPEG involves both spatial compression and temporal compression. The former is similar to JPEG, and the latter removes redundant frames.

Question 5 |

In which of the following, the density of the core remains constant from the center to the edges?

Single mode fiber | |

Multimedia step-index fiber | |

Multimode graded index fiber | |

Single mode step index fiber |

Question 5 Explanation:

Multimode Step-index: In multimode step-index fiber, the density of the core remains constant from the center to the edges. A beam of light moves through this constant density in a straight line until it reaches the interface of the core and the cladding

Question 6 |

Which of the following is a special effect in motion pictures and animation that changes one image/picture into another through a seamless transition?

Modeling | |

Morphing | |

Animating | |

Wrapping |

Question 6 Explanation:

→ Morphing is a special effect in motion pictures and animations that changes (or morphs) one image or shape into another through a seamless transition.

→ Morphing means stretching or as part of a fantasy or surreal sequence.

→ Morphing means stretching or as part of a fantasy or surreal sequence.

Question 7 |

Even when the screen is completely dark while the film is in motion, commercial motion pictures use

32 frames per second or 101 screen illuminations per second | |

72 frames per second or 234 screen illuminations per second | |

8 frames per second or 32 screen illuminations per second | |

24 frames per second or 72 screen illuminations per second |

Question 7 Explanation:

→ Motion picture, also called film or movie, series of still photographs on film, projected in rapid succession onto a screen by means of light.

→ When a motion picture film is projected on a screen at the rate of at least 16 illuminations per second but the commercial motion pictures uses 24 frames per second or 72 screen illuminations per second.

→ Film Technology standard is 24 frames per second, a three bladed shutter and some dreamy motion blur, all projected as shadow and light on the side of a wall.

→ When a motion picture film is projected on a screen at the rate of at least 16 illuminations per second but the commercial motion pictures uses 24 frames per second or 72 screen illuminations per second.

→ Film Technology standard is 24 frames per second, a three bladed shutter and some dreamy motion blur, all projected as shadow and light on the side of a wall.

Question 8 |

The file size of a 640 by 480 pictures of 256 colours in a 8-bit resolution is

300KB | |

900KB | |

128KB | |

1024KB |

Question 8 Explanation:

●Give picture size with 640 by 480 resolution with 256 colors per pixel.

●This implies 640 * 480 = 307,200 pixels. Recall that 256 values requires 8 bits of computer storage.

●Hence (307,200 pixels) * (8 bits per pixel) = 2,457,600 bits of storage.

●A common method for selling computer storage is in 8 bit groupings called bytes.

●Hence the picture would require (2,457,600 bits) / (8 bits per byte) = 307,200 bytes of computer storage.

●This implies 640 * 480 = 307,200 pixels. Recall that 256 values requires 8 bits of computer storage.

●Hence (307,200 pixels) * (8 bits per pixel) = 2,457,600 bits of storage.

●A common method for selling computer storage is in 8 bit groupings called bytes.

●Hence the picture would require (2,457,600 bits) / (8 bits per byte) = 307,200 bytes of computer storage.

Question 9 |

The process of producing bitmapped images from a view of 3-D models in a 3-D scene is called

Rendering | |

Looping | |

Cross dissolving | |

Imaging |

Question 9 Explanation:

Rendering or image synthesis is the automatic process of generating a photorealistic or non-photorealistic image from a 2D or 3D model (or models in what collectively could be called a scene file) by means of computer programs.

Question 10 |

Which of the following is interactive?

A radio broadcast | |

A talk show on TV | |

A newspaper | |

A computer game |

Question 10 Explanation:

Options A,B and C are not interactive , it will provide only information to the people. Where as Computer game in which computer user should interact with the computer in order to play the game

Question 11 |

Anti aliasing is important to improve readability of text. It deals with the

Elimination of "jaggies" | |

Spacing between two individual characters | |

Underlining letters | |

Spacing of a group of characters |

Question 11 Explanation:

● Antialiasing removes jagged edges by adding subtle color changes around the lines, tricking the human eye into thinking that the lines are not jagged.

● The slight changes in color around the edges of an image help the line blend around curves, giving the impression that the line is true.

● These color changes are made on a very small scale that the human eye cannot detect under normal circumstances. In order to be able to see that an image has been antialiased, it would have to be magnified.

● The slight changes in color around the edges of an image help the line blend around curves, giving the impression that the line is true.

● These color changes are made on a very small scale that the human eye cannot detect under normal circumstances. In order to be able to see that an image has been antialiased, it would have to be magnified.

Question 12 |

What is the bitrate for transmitting uncompressed 800x600 pixel color frame with 8 bits/pixel at 40 frames/Second?

1536 Mbps | |

2.4 Mbps | |

15.36 Mbps | |

153.6 Mbps |

Question 12 Explanation:

Step-1: Given data, Uncompressed pixel = 800*600

Step-2: Each of 8 bit = 800*600 *8

Step-3: Bit rate for transmitting = 800*600 *8 *40 =153600000 bits

Step-4: Here, they are given in Mbps. 153600000 bits =153.6Mbps

Step-2: Each of 8 bit = 800*600 *8

Step-3: Bit rate for transmitting = 800*600 *8 *40 =153600000 bits

Step-4: Here, they are given in Mbps. 153600000 bits =153.6Mbps

Question 13 |

The easiest method in flash to draw a heptagon, is to use the___

Polystar tool with the "Polygon" style | |

Polygon Tool | |

Lasso Tool with the "create star" option | |

Polystar Tool with the "star" style |

Question 13 Explanation:

There are 3 tools for drawing geometric shapes

1. Rectangle(Rectangle tool)

2. Oval(Oval tool)

3. Polygon(Polystar tool)

Note: We are not using Option A,C and D for drawing heptagon.

1. Rectangle(Rectangle tool)

2. Oval(Oval tool)

3. Polygon(Polystar tool)

Note: We are not using Option A,C and D for drawing heptagon.

Question 14 |

___ is a special effect in motion pictures and animations that changes one image or sharp into another through a seamless transition

Tweening | |

Inverse kinematics | |

Morphing | |

Tweaking |

Question 14 Explanation:

Morphing is a special effect in motion pictures and animations that changes (or morphs) one image or shape into another through a seamless transition. morphing means stretching or as part of a fantasy or surreal sequence. Traditionally such a depiction would be achieved through cross-fading techniques on film.

Question 15 |

Given that a 22 inch monitor with an aspect ratio of 16:9 has a monitor of 1920x1080, what is the width of the monitor?

22 inches | |

8:53 inches | |

10:79 inches | |

19:17 inches |

Question 15 Explanation:

Step-1: To check the ratio [width/height= 1920 /1080 ]

Step-2: Ratio: 16/9.

Step-3: Using Pythagoras theorem (9x)

^{2} +(16x)

^{ 2} =(22)

^{ 2}

x=19.17 inch

Question 16 |

Consider the following statements:

a) 3D Studio Max includes a number of high-level professional tools for character animation, game development, and visual effect production.

b) MAYA is a complete modeling package developed by Microsoft

c) RenderMan is a rendering package developed by Pixar

Which of the above statements are true?

a) 3D Studio Max includes a number of high-level professional tools for character animation, game development, and visual effect production.

b) MAYA is a complete modeling package developed by Microsoft

c) RenderMan is a rendering package developed by Pixar

Which of the above statements are true?

Only a and b | |

Only a and c | |

Only b and c | |

a,b and c |

Question 16 Explanation:

MAYA 3D animation software offers a comprehensive creative feature set for 3D computer animation, modeling, simulation, rendering, and compositing on a highly extensible production platform. Maya has next-generation display technology, accelerated modeling workflows, and tools for handling complex data.

Question 17 |

The ____ file format allows storing an animation sequence

PNG | |

GIF | |

JPG | |

PDF |

Question 17 Explanation:

→PDF means portable document format.

→Portable Network Graphics is a raster-graphics file-format that supports lossless data compression.

→JPG/JPEG: Joint Photographic Experts Group. JPEG is usually known as JPG. It stands for Joint Photographic Expert Group, a joint working group of International Standardization Organization (ISO) and International Electrotechnical Commission (IEC). It is a standard method of compressing graphic images.

→A GIF is a computer file that is used on the internet for sending images, especially moving images. GIF is an abbreviation for 'Graphic Interchange Format'.

→Portable Network Graphics is a raster-graphics file-format that supports lossless data compression.

→JPG/JPEG: Joint Photographic Experts Group. JPEG is usually known as JPG. It stands for Joint Photographic Expert Group, a joint working group of International Standardization Organization (ISO) and International Electrotechnical Commission (IEC). It is a standard method of compressing graphic images.

→A GIF is a computer file that is used on the internet for sending images, especially moving images. GIF is an abbreviation for 'Graphic Interchange Format'.

Question 18 |

If the frame buffer has 10-bits per pixel and 8-bits are allocated for each of the R,G and B components then what would be the size of the color lookup table(LUT)

(2 ^{10} +2^{ 11} ) bytes | |

(2 ^{10} +2^{ 8} ) bytes | |

(2 ^{10} + 2^{ 24} ) bytes | |

(2 ^{8} + 2^{ 9} ) bytes |

Question 18 Explanation:

10-bits per pixel it means we will have 2

8-bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24-bits( 8-bits for each of the R, G and B component).

24-bits = 3 Bytes

So the size of lookup table is = Number of entries * size of each entry

So the size of lookup table is = (2

the size of lookup table is = 3072 Bytes = (2

^{ 10} entries in color lookup table.8-bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24-bits( 8-bits for each of the R, G and B component).

24-bits = 3 Bytes

So the size of lookup table is = Number of entries * size of each entry

So the size of lookup table is = (2

^{ 10} * 3) Bytesthe size of lookup table is = 3072 Bytes = (2

^{10} + 2^{ 11} ) BytesQuestion 19 |

Which homogeneous 2D matrix transforms the figure (a) on the left side to the figure (b) on the right ?

Question 19 Explanation:

The homogeneous coordinates of the cartesian point (x,y) is (x,y,1) and the transformation Matrix is a 3x3 Matrix.

The image in figure a is translated(by 1 unit in x axis), scaled( 2 units in y axis) and rotated(90 degrees counterclockwise) to figure in b.

The image in figure a is translated(by 1 unit in x axis), scaled( 2 units in y axis) and rotated(90 degrees counterclockwise) to figure in b.

Question 20 |

In 3D Graphics, which of the following statements about perspective and parallel projection is/are true?

P: In a perspective projection, the farthest an object is from the center of projection, the smaller it appears.

Q: Parallel projection is equivalent to a perspective projection where the viewer is standing infinitely far away

R: Perspective projections do not preserve straight lines.

P: In a perspective projection, the farthest an object is from the center of projection, the smaller it appears.

Q: Parallel projection is equivalent to a perspective projection where the viewer is standing infinitely far away

R: Perspective projections do not preserve straight lines.

P and R only | |

P,Q and R | |

Q and R only | |

P and Q only |

Question 20 Explanation:

Perspective Projection :

Perspective projection is representing or drawing objects which resemble the real thing.

Perspective projection preserves the straight line.

In perspective projection, objects that are far away appear smaller, and objects that are near appear bigger.

Parallel lines do not remain parallel

Distance and angles are not preserved.

Parallel Projection :

In this projection drawing objects looks less realistic.

In this projection parallel lines remains parallel.

Angles are not preserved in this projection.

It is good for exact measurements.

Perspective projection is representing or drawing objects which resemble the real thing.

Perspective projection preserves the straight line.

In perspective projection, objects that are far away appear smaller, and objects that are near appear bigger.

Parallel lines do not remain parallel

Distance and angles are not preserved.

Parallel Projection :

In this projection drawing objects looks less realistic.

In this projection parallel lines remains parallel.

Angles are not preserved in this projection.

It is good for exact measurements.

Question 21 |

In 3D Graphics, which of the following statement/s is/are true ?

P: Back-face culling is an example of an image-precision visible-surface determination.

Q: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.

P: Back-face culling is an example of an image-precision visible-surface determination.

Q: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.

P only | |

Q only | |

Neither P nor Q | |

P and Q |

Question 21 Explanation:

Back Face Culling:

→ Back-face culling (an object space algorithm) works on 'solid' objects which you are looking at from the outside. That is, the polygons of the surface of the object completely enclose the object.

→ Back-face culling is not an example of an image-precision visible-surface determination.

→ Back-face culling can very quickly remove unnecessary polygons. Unfortunately there are often times when back-face culling can not be used. For example if you wish to make an open-topped box - the inside and the outside of the box both need to be visible, so either two sets of polygons must be generated, one set facing out and another facing in, or back-face culling must be turned off to draw that object.

1. Back faces: faces of opaque object which are “pointing away” from viewer

2. Back face culling – remove back faces (supported by OpenGL)

→ TRUE: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.

→ Back-face culling (an object space algorithm) works on 'solid' objects which you are looking at from the outside. That is, the polygons of the surface of the object completely enclose the object.

→ Back-face culling is not an example of an image-precision visible-surface determination.

→ Back-face culling can very quickly remove unnecessary polygons. Unfortunately there are often times when back-face culling can not be used. For example if you wish to make an open-topped box - the inside and the outside of the box both need to be visible, so either two sets of polygons must be generated, one set facing out and another facing in, or back-face culling must be turned off to draw that object.

1. Back faces: faces of opaque object which are “pointing away” from viewer

2. Back face culling – remove back faces (supported by OpenGL)

→ TRUE: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.

Question 22 |

Which API is used to draw a circle ?

Circle( ) | |

Ellipse( ) | |

Round Rect( ) | |

Pie( ) |

Question 22 Explanation:

→ When we are giving same height and width to a ellipse it will become circle.

→ API is a protocol intended to be used as an interface by software components to communicate with each other.

→ An API is a library that may include specification for routines, data structures, object classes and variables

→ API is a protocol intended to be used as an interface by software components to communicate with each other.

→ An API is a library that may include specification for routines, data structures, object classes and variables

Question 23 |

The LED’s for their display require

A voltage of 1.2v and a current of 20 mA | |

A voltage of 20v and a current of 1.2 A | |

A voltage of 1.2v and current of 100 mA | |

A voltage of 10v and a current of 120 mA |

Question 24 |

Two equal voltages os same frequency applied to the X and Y plates of a CRD, produces a circle on the screen. The phase difference between the two voltages is

30 ^{o} | |

90 ^{o} | |

180 ^{o} | |

0 ^{o} |

Question 25 |

The spacing between character pairs is called :

Kerning | |

x-height | |

intercap | |

serif |

Question 25 Explanation:

→ kerning is the process of adjusting the spacing between characters in a proportional font, usually to achieve a visually pleasing result.

→ Kerning adjusts the space between individual letter forms, while tracking (letter-spacing) adjusts spacing uniformly over a range of characters.

→ Kerning adjusts the space between individual letter forms, while tracking (letter-spacing) adjusts spacing uniformly over a range of characters.

Question 26 |

Which of the following is not a component of Memory tube display ?

Flooding gun | |

Collector | |

Ground | |

Liquid Crystal |

Question 26 Explanation:

Memory tube display components:

1. Connector Pins

2. Electron Gun

3. Base

4. Focusing System

5. Control Grid Voltage

6. X/Y Deflect

7. Phosphor

8. Collector

9. Ground

Note: Liquid crystal is not a component of memory tube display

.

1. Connector Pins

2. Electron Gun

3. Base

4. Focusing System

5. Control Grid Voltage

6. X/Y Deflect

7. Phosphor

8. Collector

9. Ground

Note: Liquid crystal is not a component of memory tube display

.

Question 27 |

Which of the following is not true in case of Oblique Projections?

Parallel projection rays are not perpendicular to the viewing plane. | |

Parallel lines in space appear parallel on the final projected image. | |

Used exclusively for pictorial purposes rather than formal working drawings. | |

Projectors are always perpendicular to the plane of projection. |

Question 28 |

With respect to CRT, the horizontal retrace is defined as:

The path an electron beam takes when returning to the left side of the CRT. | |

The path an electron beam takes when returning to the right side of the CRT. | |

The technique of turning the electron beam off while retracing. | |

The technique of turning the electron beam on/off while retracing. |

Question 29 |

A 4*4 DFT matrix is given by :

Where values of x and y are _____, _____ respectively.

Where values of x and y are _____, _____ respectively.

1, −1 | |

−1, 1 | |

−j, j | |

j, −j |

Question 30 |

Which of the following statement(s) is/are correct ?

Persistence is the term used to describe the duration of phosphorescence. | |

The control electrode is used to turn the electron beam on and off. | |

The electron gun creates a source of electrons which are focused into a narrow beam directed at the face of CRT. | |

All of the above |

Question 30 Explanation:

TRUE: Persistence is the term used to describe the duration of phosphorescence. Phosphorescence is a process in which energy absorbed by a substance is released relatively slowly in the form of light.

TRUE: The control electrode is used to turn the electron beam on and off.

TRUE: The electron gun creates a source of electrons which are focused into a narrow beam directed at the face of CRT.

TRUE: The control electrode is used to turn the electron beam on and off.

TRUE: The electron gun creates a source of electrons which are focused into a narrow beam directed at the face of CRT.

Question 31 |

A segment is any object described by GKS commands and data that start with CREATE SEGMENT and Terminates with CLOSE SEGMENT command. What functions can be performed on these segments ?

Translation and Rotation | |

Panning and Zooming | |

Scaling and Shearing | |

Translation, Rotation, Panning and Zooming |

Question 32 |

Match the following:

a-i, b-ii, c-iii, d-i | |

a-i, b-iii, c-ii, d-iv | |

a-iv, b-iii, c-ii, d-i | |

a-iv, b-ii, c-i, d-iii |

Question 33 |

Below are the few steps given for scan-converting a circle using Bresenham’s Algorithm. Which of the given steps is not correct ?

Compute d = 3 – 2r (where r is radius) | |

Stop if x > y | |

If d < 0, then d = 4x + 6 and x = x + 1 | |

If d ≥, then d = 4 * (x – y) + 10, x = x + 1 and y = y + 1 |

Question 33 Explanation:

__Scan converting a circle using Bresenham’s algorithm:__

1 Initially X = 0 , Y = R and D = 3 – 2R

2. While (X < Y)

3. Call Draw Circle(X

_{c}, Y

_{c}, X, Y)

4. Set X = X + 1

5. If (D < 0) Then

6. D = D + 4X + 6

7. Else

8. Set Y = Y – 1

9. D = D + 4(X – Y) + 10

10. End If

11. Draw Circle(X

_{c}, Y

_{c}, X, Y) /* calling function call */

12. End While

Question 34 |

Which of the following is/are side effects of scan conversion ?

a. Aliasing

b. Unequal intensity of diagonal lines

c. Overstriking in photographic applications

d. Local or Global aliasing

a. Aliasing

b. Unequal intensity of diagonal lines

c. Overstriking in photographic applications

d. Local or Global aliasing

a and b | |

a, b and c | |

a, c and d | |

a, b, c and d |

Question 34 Explanation:

__Side effects of scan conversions are__1. Aliasing

2. Unequal intensity of diagonal lines

3. Overstriking in photographic applications

4. Local or Global aliasing

Question 35 |

Consider a line AB with A=(0,0) and B=(8,4). Apply a simple DDA algorithm and compute the first four plots on this line.

[(0, 0), (1, 1), (2, 1), (3, 2)] | |

[(0, 0), (1, 1.5), (2, 2), (3, 3)] | |

[(0, 0), (1, 1), (2. 2.5), (3, 3)] | |

[(0, 0), (1, 2), (2, 2), (3, 2)] |

Question 35 Explanation:

Question 36 |

Consider the Breshenham’s circle generation algorithm for plotting a circle with centre (0, 0) and radius ‘r’ units in first quadrant. If the current point is (x

_{i}, y_{i}) and decision parameter is p_{i}then what will be the next point (x_{i}+ 1, y_{i}+ 1 + 1) and updated decision parameter p_{i}+ 1 for p_{i}≥ 0?x _{i} + 1 = x_{i}+ 1 y _{i} + 1 = y_{i} p _{i} + 1 = p_{i} + 4x_{i} + 6 | |

x _{i} + 1 = x_{i} + 1 y _{i} + 1 = y_{i} - 1 p _{i} + 1 = p_{i} + 4(x_{i} - y_{i}) + 10 | |

x _{i} + 1 = x_{i} y _{i} + 1 = y_{i} - 1 p _{i} + 1 = p_{i} + 4(x_{i} - y_{i}) + 6 | |

x _{i} + 1 = x_{i} - 1 y _{i} + 1 = y_{i} p _{i} + 1 = p_{i} + 4(x_{i} - y_{i}) + 10 |

Question 36 Explanation:

Write the steps required to scan - convert a circle using Bresenham’s algorithm.

Set the initial values of the variables: (h, k) = coordinates of circle center; x=0; y=circle radius r and d = 3 - 2r.

Test to determine whether the entire circle has been scan-converted. If x>y, stop.

Plot the eight points, found by symmetry with respect to the center (h, k), at the current (x, y) coordinates:

Plot(x+h, y+k) Plot(-x+h, -y+k)

Plot(y+h, x+k) Plot(-y+h, -x+k)

Plot(-y+h, x+k) Plot(y+h, -x+k)

Plot(-x+h, y+k) Plot(x+h, -y+k)

Compute the location of the next pixel. If d<0, then d=d+4x+6 and x=x+1. If d≥0, then d=d+4(x-y)+10, x=x+1 and y=y-1.

Go to step 2.

Set the initial values of the variables: (h, k) = coordinates of circle center; x=0; y=circle radius r and d = 3 - 2r.

Test to determine whether the entire circle has been scan-converted. If x>y, stop.

Plot the eight points, found by symmetry with respect to the center (h, k), at the current (x, y) coordinates:

Plot(x+h, y+k) Plot(-x+h, -y+k)

Plot(y+h, x+k) Plot(-y+h, -x+k)

Plot(-y+h, x+k) Plot(y+h, -x+k)

Plot(-x+h, y+k) Plot(x+h, -y+k)

Compute the location of the next pixel. If d<0, then d=d+4x+6 and x=x+1. If d≥0, then d=d+4(x-y)+10, x=x+1 and y=y-1.

Go to step 2.

Question 37 |

A point P(5, 1) is rotated by 90° about a pivot point (2, 2). What is the coordinate of new transformed point P′ ?

(3, 5) | |

(5, 3) | |

(2, 4) | |

(1, 5) |

Question 37 Explanation:

Rotation around a pivot point (2, 2) can be represented as:

(h, k) = (2, 2)

(h, k) = (2, 2)

Question 38 |

Let R be the rectangular window against which the lines are to be clipped using 2D Sutherland-Cohen line clipping algorithm. The rectangular window has a lower left-hand corner at (– 5, 1) and upper right-hand corner at (3, 7). Consider the following three lines for clipping with the given endpoint coordinates:

Line AB : A (– 6, 2) and B (–1, 8)

Line CD : C (– 1, 5) and D (4, 8)

Line EF : E (–2, 3) and F (1, 2)

Which of the following line(s) is/are candidate for clipping?

Line AB : A (– 6, 2) and B (–1, 8)

Line CD : C (– 1, 5) and D (4, 8)

Line EF : E (–2, 3) and F (1, 2)

Which of the following line(s) is/are candidate for clipping?

AB | |

CD | |

EF | |

AB and CD |

Question 39 |

In perspective projection, if a line segment joining a point which lies in front of the viewer to a point in back of the viewer is projected to a broken line of infinite extent. This is known as _______.

View confusion | |

Vanishing point | |

Topological distortion | |

Perspective foreshortening |

Question 40 |

Let us consider that the original point is (x, y) and new transformed point is (x′, y′). Further, Shx and Shy are shearing factors in the x and y directions. If we perform the y-direction shear relative to x = xref then the transformed point is given by _______.

x′ = x + Shx ⋅ (y – y _{ref})y′ = y | |

x′ = x y′ = y ⋅ Sh _{x} | |

x′ = x y′ = Sh _{y} (x – x_{ref}) + y | |

x′ = Sh _{y} ⋅ y y′ = y ⋅ (x – xref) |

Question 40 Explanation:

Question 41 |

Which of the following statement(s) is/are correct with reference to curve generation?

I. Hermite curves are generated using the concepts of interpolation.

II. Bezier curves are generated using the concepts of approximation.

III. The Bezier curve lies entirely within the convex hull of its control points.

IV. The degree of Bezier curve does not depend on the number of control points.

I. Hermite curves are generated using the concepts of interpolation.

II. Bezier curves are generated using the concepts of approximation.

III. The Bezier curve lies entirely within the convex hull of its control points.

IV. The degree of Bezier curve does not depend on the number of control points.

I, II and IV only | |

II and III only | |

I and II only | |

I, II and III only |

Question 42 |

A triangulation of a polygon is a set of T chords that divide the polygon into disjoint triangles. Every triangulation of n-vertex convex polygon has _____ chords and divides the polygon into _____ triangles.

n – 2, n – 1 | |

n – 3, n – 2 | |

n – 1, n | |

n – 2, n – 2 |

Question 43 |

Consider a raster grid having XY-axes in positive X-direction and positive upward Y-direction with X

_{max}= 10, X_{min}= –5, Y_{max}= 11, and Y_{min}= 6. What is the address of memory pixel with location (5, 4) in raster grid assuming base address 1 (one) ?150 | |

151 | |

160 | |

161 |

Question 44 |

Consider a N-bit plane frame buffer with W-bit wide lookup table with W > N. How many intensity levels are available at a time ?

2 ^{N} | |

2 ^{W} | |

2 ^{N+W } | |

2 ^{ N-1} |

Question 45 |

Consider the Bresenham's line generation algorithm for a line with gradient greater than one, current point (x

_{i}, y_{i}) and decision parameter, d_{i}. The next point to be plotted (x_{i+1}, y_{i+1}) and updated decision parameter, d_{i+1}, for d_{i}< 0 are given as _______.x _{i+1} = x_{i +1} y _{i+1} = y_{i} d _{i+1} = d_{i+ 2} dy | |

x _{i+1} = x_{i} y _{i+1} = y_{i} +1 d _{i+1} = d_{i}+ 2 dx | |

x _{i+1} = x_{i }y _{i+1} = y_{i} +1 d _{i+1} = d_{i}+ 2(dx -dy) | |

x _{i+1} = x_{i }+1y _{i+1} = y_{i }+1 d _{i+1} = d_{i}+ 2(dy -dx) |

Question 45 Explanation:

Question 46 |

A point P(2, 5) is rotated about a pivot point (1, 2) by 60°. What is the new transformed point P' ?

(1, 4) | |

(–1, 4) | |

(1, – 4) | |

(– 4, 1) |

Question 46 Explanation:

Question 47 |

In perspective projection (from 3D to 2D), objects behind the centre of projection are projected upside down and backward onto the view-plane. This is known as _____.

Topological distortion | |

Vanishing point | |

View confusion | |

Perspective foreshortening |

Question 48 |

The Liang-Barsky line clipping algorithm uses the parametric equation of a line from (x

x = x

y = y

Where ∆x = x

The lower and upper bound values of the parameter u for the clipped line using Liang-Barsky algorithm is given as :

_{1}, y_{1}) to (x_{2}, y_{2}) along with its infinite extension which is given as :x = x

_{1}+ ∆x.uy = y

_{1}+ ∆y.uWhere ∆x = x

_{2}– x_{1}, ∆y = y_{2}– y_{1}, and u is the parameter with 0 ≤ u ≤ 1. A line AB with endpoints A(–1, 7) and B(11, 1) is to be clipped against a rectangular window with x_{min}= 1, x_{max}= 9, y_{min}= 2, and y_{max}= 8.The lower and upper bound values of the parameter u for the clipped line using Liang-Barsky algorithm is given as :

(0, 2/3) | |

(1/6 , 5/6) | |

(0, 1/3) | |

(0, 1) |

Question 49 |

Which of the following graphic primitives are considered as the basic building blocks of computer graphics ?

(a)Points

(b)Lines

(c)Polylines

(d)Polygons

(a)Points

(b)Lines

(c)Polylines

(d)Polygons

(a) only | |

(a) and (b) | |

(a), (b) and (c) | |

(a), (b), (c) and (d) |

Question 49 Explanation:

Ponits, lines and polygon lines are the basic building blocks of computer graphics

Question 50 |

Which of the following is/are the principal components of a memory-tube display ?

(a)Flooding gun

(b)Collector

(c)Phosphorus grains

(d)Ground

(a)Flooding gun

(b)Collector

(c)Phosphorus grains

(d)Ground

(a) and (b) | |

(c) only | |

(d) only | |

All the above |

Question 50 Explanation:

Question 51 |

Which raster locations would be chosen by Bresenham’s algorithm when scan converting a line from (1, 1) to (8, 5)?

Question 51 Explanation:

Question 52 |

Consider a unit square centred at the origin. The coordinates of the square are translated by a factor (1/2, 1) and rotated by an angle of 90°. What shall be the coordinates of the new square?

Question 52 Explanation:

Question 53 |

Which of the following is/are the components of a CRT?

(a)Electron Gun

(b)Control Electrode

(c)Focusing Electrode

(d)Phosphor Coated Screen

(a)Electron Gun

(b)Control Electrode

(c)Focusing Electrode

(d)Phosphor Coated Screen

(a) and (d) | |

(a), (b) and (d) | |

(a), (b), (c) and (d) | |

(a), (c) and (d) |

Question 53 Explanation:

Question 54 |

Using the phong reflectance model, the strength of the specular highlight is determined by the angle between

The view vector and the normal vector | |

The light vector and the normal vector | |

The light vector and the reflected vector
| |

the reflected vector and the view vector |

Question 54 Explanation:

Specular highlights on objects appear, because the light reflected from the object hits viewer. Sometimes, we are hit directly by the light and the highlight appears to be very intense, and sometimes we are hit so slightly, that we basically don't see the specular highlight. The strength of this highlight depends on the cosine of the angle between the reflected vector and vector taken from the point at object to the viewer's position:

Question 55 |

Give the number of principal vanishing point(s) along with their direction for the standard perspective transformation:

Only one in the direction K | |

Two in the directions I and J | |

Three in the directions I, J and K | |

Only two in the directions J and K |

Question 55 Explanation:

There is only one principal vanishing point and it is in K-direction.

Question 56 |

(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv), (e)-(v) | |

(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv), (e)-(v) | |

(a)-(iii), (b)-(i), (c)-(ii), (d)-(v), (e)-(iv) | |

(a)-(iv), (b)-(v), (c)-(i), (d)-(ii), (e)-(iii) |

Question 56 Explanation:

Flood Gun: An electron gun designed to flood the entire screen with
electrons.

Collector: Partly energized by flooding gun stores the charge generated by the writing gun.

Ground: Used to discharge the collector.

Collector: Partly energized by flooding gun stores the charge generated by the writing gun.

Ground: Used to discharge the collector.

Question 57 |

If we want to resize a 1024 × 768 pixels image to one that is 640 pixels wide with the same aspect ratio, what would be the height of the resized image?

420 Pixels | |

460 Pixels | |

480 Pixels | |

540 Pixels |

Question 57 Explanation:

Aspect Ratio= Width / Height

Aspect ration of 1024 × 768 pixels image = 1024/768

= 4/3

Aspect ration of modified pixels image = 640/height

4/3 = 640/height

Height = (3*640)/4

Height = 480 Pixels

Aspect ration of 1024 × 768 pixels image = 1024/768

= 4/3

Aspect ration of modified pixels image = 640/height

4/3 = 640/height

Height = (3*640)/4

Height = 480 Pixels

Question 58 |

Consider the following statement with respect to approaches to fill area on raster systems:

(P) To determine the overlap intervals for scan lines that cross the area.

(Q) To start from a given interior position and paint outward from this point until we encounter the specified boundary conditions.

Select the correct answer from the options given below:

(P) To determine the overlap intervals for scan lines that cross the area.

(Q) To start from a given interior position and paint outward from this point until we encounter the specified boundary conditions.

Select the correct answer from the options given below:

P only | |

Q only | |

Both P and Q | |

Neither P nor Q |

Question 58 Explanation:

Fill area on raster systems:

TRUE: To determine the overlap intervals for scan lines that cross the area.

TRUE: To start from a given interior position and paint outward from this point until we encounter the specified boundary conditions.

TRUE: To determine the overlap intervals for scan lines that cross the area.

TRUE: To start from a given interior position and paint outward from this point until we encounter the specified boundary conditions.

There are 58 questions to complete.