CS 39J: Session 12

http://inst.eecs.berkeley.edu/~cs39j/session12.html
18 April 2002

Photoshop

With a live computer demonstration, Kawaldeep explained the use of masks, selections, dodge, and burn.

Suggestions for next time

We discuss some suggestions for future CS 39J sessions and classes. Some of them include:

• Have more lectures concentrating on different photographers and the historical background of photography.
• Photography was not considered an art form awhiles back. Nowadays it's considered an art form. The good news is that people appreciate photography more. The bad news is that photography critics are generally not aware about the technical aspects of photography.
• Prevalent themes in photography; perhaps trends in technical aspects or styles in photography.

Dr. Stan Klein Dr. Stan Klein is a professor from the school of optometry, and gave a special lecture on the science behind optics.

On Lenses

"Quality lenses" are more expensive because more is done to eradicate aberrations on the lens's surface. The most dramatic, visible aberration you can see in a lens is curvature of the field, or defocus. There's also distortion, spherical aberration, and chromatic aberration (where the different colors of light separate). These aberrations can be fixed by combining different glasses made from different materials and curvatures. Usually, this is figured out by sticking the data into a computer.

1) What is the relationship between the object and the lens?

In the history of mankind, there are two ways people have described nature: through local and through non-local effects. Modern science has replaced nonlocal effects with local ones by, for example, inventing fields to explain gravity and electricity. The relationship of object and image is similar. One could connect the two in a nonlocal way by relating the distance of the object and image 1/u - 1/v = 1/f where u and v are the locations of the object and image using the lens as the origin of the coordinate system and f is the focal length. This is a nonlocal looking equation since the location of the image is distant from the location of the object. One could rewrite this as a local equation by writing it in terms of the curvatures of the wave entering and leaving the lens:

D = U - V

where U=1/u, V=1/v and D=1/f is the change in diopters of curvature of the wave.

The lens only knows the curvature of the waves from an object's light. The lens only changes the curvature of the light. Curvature is the inverse of a distance, measured as C = 1/distance. The smaller the radius, the more curved it is.

Photographers are generally interested in two problems:

• Field of view. For example, if you use a telephoto lens, how wide is your view?
• How do you calculate depth-of-field? For example, you might want to take a photo of a pretty flower but make the surrounding background blurred.

Depth-of-field

Depth-of-focus is "how far can I focus the film and not be bothered?" Depth-of-field is "how far can move my object and still have it in tolerable focus?" I'm going to discuss this criterion as an angle. The angle I'm interested in is , the blur angle, which we discuss in radians. The blur is calculated by the formula:

= p * D (Eq. 1)

where p is the size of the lens aperture that I'll call the pupil, and D = "delta diopters" is the change in curvature of caused by the change in object location (for depth of field) or image location (depth of focus). I want to be 1 minute, or (20/20). In radians this angle is =1 (min)*1/60 (deg/min)*1/57.3 (rad/deg) = 1/3438 radians. This is the smallest unit in which the human eye can distinguish a blurred and a sharp object. Each pixel is about 1/2 mm on a typical TV. The blur is thus 1/2 mm. For a TV's pixel, theta is 1/2000 if the TV is 1000 mm in front of the eye.

...

The bigger the pupil, the bigger the aperture, and what you see gets blurry. Thus, if you squint your eyes, what you see becomes sharper. In a camera the pupil size is given by the focal length divided by the f-number. Thus a 50 mm, f/5.6 lens would have a pupil size of about p = 0.05/5.6 = .009 m. Thus from Eq. 1 the tolerable delta diopters is:

D = theta/p = .0005/.009 = .06 diopters.

If the object is 2 meters away the depth of field would be:

1/d - 1/d0 = D so d = d0/(1+d0*D) = 2/(1+2*.06) = 1.78 m

That is, moving the object from 2 m to 1.78 m would just cause a point image on one pixel to substantially blur into the next pixel.

Field of view

The field-of-view is also very simple. The field-of-view is the aspect ratio, which is the ratio of image size to the image distance from lens, which is the focal length. For a standard 35 mm camera with a 50 mm focal length, this aspect ratio is R = 35/50 = 0.7. Since light going through a camera does not bend, the image space of the field-of-view is the same as the object field-of-view. It's as simple as that.

The wide-angle lens has a shorter focal length, and thus a bigger field-of-view.

For digital cameras, the CCD array (that captures the light for digital cameras) is not 35 mm like traditional film, but instead around 20 mm. To get the same field-of-view, one would need a shorter focal length. It gets messy with the lenses' nomenclature.

24 x 36 is the aspect ratio of 35 mm film. Some of the digital cameras do not have the same aspect ratio as that; they tend to be less wide and more like a computer monitor.

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Current webmaster: Steven Chan (mr_chan@uclink.berkeley.edu). CS 39J: The Art and Science of Photography is a freshman seminar taught by Professor Brian A. Barsky. Site design by Steven Chan.