Top 30 Computer Vision Interview Questions and Answers
30 multiple-choice questions on Computer Vision, of the kind that come up in a technical interview, drawn from 30 bites in the Computer Vision library. Answer them here or read straight down. Every question carries the correct option, why it is correct, and a link to the bite it came from.
Image/video models, diffusion, OCR, multimodal
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Question 1 of 30
Why is HSV often preferred over RGB for segmenting a colored object across a scene with mixed sunlight and shadow?
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Answer: b · Hue stays relatively stable under lighting changes, so color separates from brightness
HSV isolates hue from value, so a color keeps its hue under varying illumination, easing thresholding. Both spaces cover the same color gamut, so the capacity claim is false.
Question 2 of 30
A photography platform archives original images for future editing and serves compressed previews. Which strategy best preserves fidelity while optimizing delivery?
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Answer: d · Store originals in a lossless format and serve previews as lossy JPEG to reduce bandwidth.
Lossless storage preserves bit-exact originals for future editing and avoids generational degradation, while lossy JPEG cuts preview sizes for faster web delivery. Option C is tempting because PNG guarantees exact pixels, but serving previews losslessly wastes bandwidth without perceptible quality gains over a well-compressed lossy image.
Read the full bite: What is the difference between lossy and lossless image compression?
Question 3 of 30
Which statement best describes how histogram equalization remaps grayscale intensities?
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Answer: d · It applies the cumulative distribution function to redistribute intensities toward a uniform histogram.
Histogram equalization uses the cumulative distribution function of the original histogram to remap intensities so the output approximates a uniform distribution, maximizing global contrast. The first option describes linear contrast stretching, which merely rescales the minimum and maximum values without considering the frequency of each intensity level.
Read the full bite: Describe a grayscale histogram and its use in exposure and equalization
Question 4 of 30
How are the intrinsic parameters fx and fy in K derived from physical camera properties?
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Answer: a · They equal the focal length divided by pixel width and height respectively
fx and fy convert the physical focal length into pixel units by dividing by pixel width and height, making C correct. A is a common misconception because K stores focal length in pixel units, not millimeters, and D is wrong since distortion is modeled outside the idealized pinhole matrix.
Read the full bite: Explain the pinhole camera model and intrinsic matrix K
Question 5 of 30
During checkerboard calibration, why must you capture many images of the board at different orientations rather than a single frontal shot?
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Answer: b · Varied views provide the geometric diversity needed to solve stably for intrinsics and distortion
Multiple poses constrain the parameter estimation enough to recover intrinsics and distortion coefficients reliably. One frontal view is degenerate and underdetermines the solution; it has nothing to do with resolution.
Question 6 of 30
Why is demosaicing necessary after a Bayer sensor captures an image?
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Answer: a · Because each photosite records only one color channel, leaving missing values to estimate.
Demosaicing is required because every photosite measures only a single color channel, so the missing two channels must be interpolated from neighbors. Option D represents the common misconception that Bayer pixels already contain complete RGB data.
Read the full bite: How does a Bayer filter capture color and what is demosaicing?
Question 7 of 30
Switching from RGB to YCbCr does not reduce uncompressed frame size, yet 4:2:0 YCbCr cuts bandwidth roughly in half. What best explains where the savings come from?
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Answer: d · Separating luma from chroma allows chroma planes to be stored at lower spatial resolution because human eyes have lower color spatial acuity.
The YCbCr transform is lossless and does not reduce uncompressed size; savings come from chroma subsampling, which exploits the human visual system's lower spatial resolution for color versus brightness. Distractor B is wrong because the transform does not inherently use fewer bits per pixel—it merely enables efficient subsampling and quantization.
Read the full bite: Compare YCbCr and RGB. Why chroma subsampling for compression?
Question 8 of 30
Why does a fixed Euclidean distance threshold in RGB fail to produce consistent perceptual segmentation across light and dark image regions?
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Answer: a · The same numerical RGB delta can correspond to a huge perceived shift in one region while being nearly invisible in another
RGB is linear with respect to light intensity but not human perception, so identical Euclidean deltas can look huge in one region and nearly invisible in another. HSV is a cylindrical transform of RGB and is not perceptually uniform, while CIELAB is designed for perceptual uniformity rather than linear physical intensity.
Read the full bite: Why is RGB Euclidean distance a poor measure of perceptual color difference?
Question 9 of 30
When photometric stereo assumes a Lambertian model but the surface is glossy, false geometry is reconstructed because the solver...
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Answer: c · misinterprets view-dependent specular brightness as a tilted surface normal
Under a Lambertian assumption, the solver expects brightness to depend only on the light direction and surface normal, so view-dependent specular highlights are misread as changes in normal orientation. Option A describes a common conceptual error—treating the BRDF as a scalar albedo—but it does not explain the specific mechanism that creates phantom geometry in photometric stereo.
Read the full bite: Describe the BRDF, its advantage over Lambertian, and critical CV tasks
Question 10 of 30
When implementing a box blur, why is it important to write results into a separate destination buffer rather than updating the source image in place?
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Answer: c · It prevents already-blurred pixel values from being reused in later neighborhood averages
Using a separate destination buffer guarantees that every neighborhood average reads only original pixel values, not values that have already been blurred and would distort subsequent averages. The overflow issue in option B is addressed by using a larger type for the accumulator during the sum, not by allocating a second image buffer.
Read the full bite: How would you implement a simple box blur on a grayscale image?
Question 11 of 30
A grayscale image looks dull because most pixels are clustered between intensity 100 and 150. After histogram equalization, what has fundamentally changed about the pixel intensities?
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Answer: c · A transfer function based on the cumulative intensity distribution was used as a lookup table to spread values across the full range.
Histogram equalization computes the cumulative distribution function from the histogram, normalizes it to the maximum intensity, and uses it as a lookup table to remap pixels across the full range. Option A describes linear contrast stretching, which only scales the min and max values without considering the actual probability distribution of intensities.
Read the full bite: What is an image histogram and how does histogram equalization improve contrast?
Question 12 of 30
Why does a median filter remove salt-and-pepper specks more cleanly than a Gaussian blur of similar size?
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Answer: c · Extreme outlier pixels sort to the ends of the window and are never chosen as the median
Median selection ignores extreme values, so corrupted black or white pixels are discarded while edges stay sharp. Gaussian blur averages those outliers into the result, smearing the noise and blurring edges.
Question 13 of 30
Which statement best explains why the Sobel Gx kernel has its specific 3x3 weight pattern?
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Answer: b · It factors into a horizontal central-difference filter and an orthogonal vertical smoothing filter.
The Sobel Gx kernel is separable into a horizontal central-difference row and a vertical smoothing column, which reduces noise while estimating the partial derivative. Distractor A is wrong because, although the smoothing weights loosely approximate a Gaussian, the kernel is separable and explicitly not rotationally invariant.
Read the full bite: How does the Sobel operator approximate image gradients for edge detection?
Question 14 of 30
When applying a 3x3 convolution to the top border of a bright photo, which padding mode preserves spatial dimensions while avoiding dark vignettes and flat streaking?
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Answer: c · Reflect-padding, because it mirrors edge pixels to assume continuity across the boundary
Reflect-padding mirrors edge pixels to maintain continuity across the boundary, avoiding both the dark vignettes caused by zero-padding and the flat streaking caused by replicate-padding. Replicate-padding is tempting because it avoids darkening, but it creates frozen-edge artifacts by repeating the same pixel value outward.
Read the full bite: Zero-padding vs reflect vs replicate padding and their visual artifacts
Question 15 of 30
When rotating an image, why is inverse mapping preferred over forward mapping of pixels?
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Answer: b · It guarantees every output pixel gets exactly one interpolated value, avoiding holes and overlaps
Iterating over output pixels and sampling the source ensures full, single-valued coverage, while forward mapping leaves gaps and collisions. Inverse mapping still uses the inverse transform and interpolation to read fractional source coordinates.
Read the full bite: Image rotation: forward versus inverse mapping
Question 16 of 30
What is the primary reason a 2D Gaussian blur can be computed in O(N²K) rather than O(N²K²) for an N×N image and K×K kernel?
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Answer: d · The Gaussian kernel is a rank-one matrix expressible as the outer product of two 1D vectors
The Gaussian kernel is separable because it equals the outer product of two 1D Gaussians (a rank-one matrix), so two O(N²K) 1D passes replace one O(N²K²) 2D convolution. Using FFT is a distinct optimization, and neither normalization nor circular symmetry implies that a kernel can be decomposed into 1D passes.
Read the full bite: How does filter separability optimize Gaussian blur and its complexity?
Question 17 of 30
Why does Canny use gradient orientation during non-maximum suppression?
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Answer: b · To determine which neighboring pixels to compare for ridge thinning
Gradient orientation tells NMS which neighbors lie along the edge direction so it can thin multi-pixel ridges to single-pixel width. Option C describes hysteresis, which uses spatial connectivity rather than gradient direction to link edges.
Read the full bite: Walk me through Canny edge detection and why it beats Sobel thresholding
Question 18 of 30
Why does the Harris detector consider a corner a more reliable tracking feature than a point along a straight edge?
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Answer: a · A corner constrains position in two directions, while an edge point can slide along the edge
At a corner, intensity changes in every shift direction, giving large eigenvalues and a precise 2D location. Along an edge only one direction constrains position, so the point slides (the aperture problem).
Read the full bite: Harris corner detector and corner stability
Question 19 of 30
How does the Canny edge detector relate to the Sobel operator in a typical pipeline?
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Answer: a · Canny uses Sobel-style gradients, then adds non-maximum suppression and hysteresis thresholding
Canny is a multi-stage pipeline that computes gradients (as Sobel does), thins them with non-maximum suppression, and links them with double-threshold hysteresis. Sobel alone gives thick, noisy edges.
Question 20 of 30
In SIFT, which step is specifically responsible for rotation invariance of the descriptor?
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Answer: c · Assigning each keypoint a dominant gradient orientation and describing it relative to that
Computing the descriptor relative to a keypoint's dominant orientation makes it invariant to rotation. Scale-space extrema give scale invariance, and normalization mainly addresses illumination, not rotation.
Question 21 of 30
For a real-time feature tracker on a battery-constrained phone, why is ORB often chosen over SIFT?
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Answer: b · ORB's binary descriptors and FAST keypoints are far cheaper to compute and match
ORB's FAST detector and short binary descriptors give real-time speed and a small footprint, ideal for mobile. SIFT is more accurate and robust but too costly; ORB trades some robustness for performance.
Question 22 of 30
Why does Lowe's ratio test reject a match whose nearest and second-nearest descriptor distances are nearly equal?
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Answer: b · Near-equal distances signal an ambiguous match with no distinctive best candidate
A reliable correspondence should be clearly closer than any alternative; when the two best are tied, the feature is not distinctive and the match is likely wrong. An absolute distance threshold would not capture this relative ambiguity.
Read the full bite: Descriptor matching and Lowe's ratio test
Question 23 of 30
What information about an image does the standard Bag of Visual Words representation deliberately discard?
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Answer: c · The spatial positions and arrangement of the features
BoVW counts word frequencies into a histogram and ignores where features appear, just like a text bag-of-words ignores word order. Vocabulary size and classifier choice are separate design decisions, not discarded image content.
Question 24 of 30
When repurposing a pretrained CNN like VGG16 for image retrieval, why are deep-layer activations usually preferred over the final softmax output?
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Answer: a · Deep activations encode rich semantic features, while softmax collapses the image to class probabilities
A late hidden layer yields a high-dimensional semantic descriptor ideal for similarity comparison, whereas the softmax discards detail by reducing the image to class scores. Earlier layers still carry useful low-level information, just less semantic content.
Question 25 of 30
Why is ORB typically preferred over SIFT for the front end of a real-time visual SLAM system on a mobile device?
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Answer: d · ORB's cheap FAST keypoints and binary descriptors meet the per-frame latency and power budget
Real-time SLAM needs detection, description, and matching within milliseconds at low power, which ORB's FAST keypoints and Hamming-matched binary descriptors satisfy. SIFT is more robust but too slow; temporal continuity offsets ORB's weaker invariance.
Read the full bite: Feature choice for real-time mobile SLAM
Question 26 of 30
Knowing point p1 in the first image, why does the epipolar constraint reduce the search for its match p2 to a single line in the second image?
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Answer: d · Because p1's 3D point lies along one viewing ray, which projects to a line in the second image
The unknown depth of p1 means its 3D point spans a ray, and that ray projects to the epipolar line in the second image where p2 must lie. Identical intrinsics or rectification are not required for the constraint to hold.
Read the full bite: Epipolar constraint for correspondence search
Question 27 of 30
In a rectified stereo pair, how does the depth of a scene point relate to its disparity?
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Answer: a · Depth is inversely proportional to disparity, given the baseline and focal length
Depth equals focal length times baseline divided by disparity, so larger disparity means a nearer point. The direct-proportion and independence options invert or ignore this geometric relationship.
Question 28 of 30
What extra information does computing the essential matrix require that the fundamental matrix does not?
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Answer: c · The camera intrinsic parameters to normalize the coordinates
The essential matrix works in calibrated normalized coordinates, so it needs the camera intrinsics, whereas the fundamental matrix is estimated from pixel correspondences alone. Absolute scale remains unknown even with the essential matrix.
Read the full bite: Fundamental matrix versus essential matrix
Question 29 of 30
In an incremental SfM pipeline, what is the primary role of bundle adjustment?
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Answer: d · To jointly refine all camera poses and 3D points by minimizing reprojection error
Bundle adjustment is the nonlinear optimization that simultaneously adjusts poses and points to reduce reprojection error and control drift. Densification is a later multi-view stereo step, and matching and seed selection happen before it.
Read the full bite: Incremental Structure from Motion pipeline
Question 30 of 30
In incremental Structure from Motion, what makes PnP the right tool for adding a new camera after the initial pair?
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Answer: c · It estimates the new camera's pose by matching its 2D features to existing 3D points
PnP recovers pose from known 3D-to-2D correspondences, registering each new frame into the existing cloud. The essential-matrix and joint-refinement options describe two-view geometry and bundle adjustment respectively, not single-camera registration.
Read the full bite: The PnP problem in Structure from Motion
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