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A worker uses a computer monitor at a viewing distance of 20 inches (a) What would be the smallest possible height of the ico

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Answer #1

a]

The distance of the retina from the cornea is approximately 1.7cm.

We know that the magnification for a lens is given by m = -v / u = h2 / h1.

Thus, m = +1.7 cm / (20 x 2.54) cm = 0.0335

=> himage / hobject = 0.0335

Now from the given data, angle subtended at retina, we can calculate himage as:

himage = (subtended angle in radians) x (image distance)

= 1.7x60 x 180

himage =  2.473 x 10-3 cm

\small \therefore hobject = himage / m = 0.074 cm = 0.74 mm

Thus the smallest possible height of the icons/characters for 20/20 vision is 0.74 mm.

b]

Since the object has moved farther away, the pupil increases its diameter in order to allow more light in. Also, the ciliary muscles contract, flattening the lens in order to focus the object which is now at a greater distance.

c]

Monocular depth cues that allow depth perception can be stated as follows:   

  • Relative heights of objects: In the picture, one of the traffic light poles seems to be at a higher elevation than the other, and hence results in us concluding that the shorter one is farther.
  • Relative size of the objects: The cars in this case. The taxi's; one seems to be much larger than the one behind(that in the middle), leading us to conclude that the smaller one is farther.
  • Parallel lines appear to meet that father they seem to go: The road edges are thus a good indicator of the depth, as they appear to converge to the centre of the image, which is thus the farther part.
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