Light enters the eye through the pupil and strikes the retina, where photoreceptor cells sense light and color. W. Stanley Stiles and B. H. Crawford studied the phenomenon in which measured brightness decreases as light enters farther from the center of the pupil. (See the figure.) A light beam A that enters through the center of the pupil measures brighter than a beam B entering near the edge of the pupil. They detailed their findings of this phenomenon, known as the Stiles–Crawford effect of the first kind , in an important paper published in 1933. In particular, they observed that the amount of luminance sensed was not proportional to the area of the pupil as they expected. The percentage P of the total luminance entering a pupil of radius r mm that is sensed at the retina can be described by P = 1 − 10 − ρ r 2 ρ r 2 ln 10 where ρ is an experimentally determined constant, typically about 0.05. (a) What is the percentage of luminance sensed by a pupil of radius 3 mm? Use ρ = 0.05. (b) Compute the percentage of luminance sensed by a pupil of radius 2 mm. Does it make sense that it is larger than the answer to part (a)? (c) Compute lim x → 0 + P . Is the result what you would expect? Is this result physically possible? Source: Adapted from W. Stiles and B. Crawford, “The Luminous Efficiency of Rays Entering the Eye Pupil at Different Points.” Proceedings of the Royal Society of London, Series B: Biological Sciences 112 (1933): 428–50.
Light enters the eye through the pupil and strikes the retina, where photoreceptor cells sense light and color. W. Stanley Stiles and B. H. Crawford studied the phenomenon in which measured brightness decreases as light enters farther from the center of the pupil. (See the figure.) A light beam A that enters through the center of the pupil measures brighter than a beam B entering near the edge of the pupil. They detailed their findings of this phenomenon, known as the Stiles–Crawford effect of the first kind , in an important paper published in 1933. In particular, they observed that the amount of luminance sensed was not proportional to the area of the pupil as they expected. The percentage P of the total luminance entering a pupil of radius r mm that is sensed at the retina can be described by P = 1 − 10 − ρ r 2 ρ r 2 ln 10 where ρ is an experimentally determined constant, typically about 0.05. (a) What is the percentage of luminance sensed by a pupil of radius 3 mm? Use ρ = 0.05. (b) Compute the percentage of luminance sensed by a pupil of radius 2 mm. Does it make sense that it is larger than the answer to part (a)? (c) Compute lim x → 0 + P . Is the result what you would expect? Is this result physically possible? Source: Adapted from W. Stiles and B. Crawford, “The Luminous Efficiency of Rays Entering the Eye Pupil at Different Points.” Proceedings of the Royal Society of London, Series B: Biological Sciences 112 (1933): 428–50.
Solution Summary: The author explains that the value of percentage of luminance sensed by a pupil of radius 3 mm is approximately 0.62%.
Light enters the eye through the pupil and strikes the retina, where photoreceptor cells sense light and color. W. Stanley Stiles and B. H. Crawford studied the phenomenon in which measured brightness decreases as light enters farther from the center of the pupil. (See the figure.)
A light beam A that enters through the center of the pupil measures brighter than a beam B entering near the edge of the pupil.
They detailed their findings of this phenomenon, known as the Stiles–Crawford effect of the first kind, in an important paper published in 1933. In particular, they observed that the amount of luminance sensed was not proportional to the area of the pupil as they expected. The percentage P of the total luminance entering a pupil of radius r mm that is sensed at the retina can be described by
P
=
1
−
10
−
ρ
r
2
ρ
r
2
ln
10
where ρ is an experimentally determined constant, typically about 0.05.
(a) What is the percentage of luminance sensed by a pupil of radius 3 mm? Use ρ = 0.05.
(b) Compute the percentage of luminance sensed by a pupil of radius 2 mm. Does it make sense that it is larger than the answer to part (a)?
(c) Compute
lim
x
→
0
+
P
. Is the result what you would expect? Is this result physically possible?
Source: Adapted from W. Stiles and B. Crawford, “The Luminous Efficiency of Rays Entering the Eye Pupil at Different Points.” Proceedings of the Royal Society of London, Series B: Biological Sciences 112 (1933): 428–50.
5. [-/1 Points]
DETAILS
MY NOTES
SESSCALCET2 6.5.AE.003.
y
y= ex²
0
Video Example
x
EXAMPLE 3
(a) Use the Midpoint Rule with n = 10 to approximate the integral
कर
L'ex²
dx.
(b) Give an upper bound for the error involved in this approximation.
SOLUTION
8+2
1
L'ex² d
(a) Since a = 0, b = 1, and n = 10, the Midpoint Rule gives the following. (Round your answer to six decimal places.)
dx Ax[f(0.05) + f(0.15) + ... + f(0.85) + f(0.95)]
0.1 [0.0025 +0.0225
+
+ e0.0625 + 0.1225
e0.3025 + e0.4225
+ e0.2025
+
+ e0.5625 €0.7225 +0.9025]
The figure illustrates this approximation.
(b) Since f(x) = ex², we have f'(x)
=
0 ≤ f'(x) =
< 6e.
ASK YOUR TEACHER
and f'(x) =
Also, since 0 ≤ x ≤ 1 we have x² ≤
and so
Taking K = 6e, a = 0, b = 1, and n = 10 in the error estimate, we see that an upper bound for the error is as follows. (Round your final
answer to five decimal places.)
6e(1)3
e
24(
=
≈
2. [-/1 Points]
DETAILS
MY NOTES
SESSCALCET2 6.5.015.
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.)
ASK YOUR TEACHER
3
1
3 +
dy, n = 6
(a) the Trapezoidal Rule
(b) the Midpoint Rule
(c) Simpson's Rule
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This question builds on an earlier problem. The randomized numbers may have changed, but have your work for the previous problem available to help with this one.
A 4-centimeter rod is attached at one end to a point A rotating counterclockwise on a wheel of radius 2 cm. The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel. At time t=0 the rod is situated as in the diagram at the left below. The
wheel rotates counterclockwise at 1.5 rev/sec. At some point, the rod will be tangent to the circle as shown in the third picture.
B
A
B
at some instant, the piston will be tangent to the circle
(a) Express the x and y coordinates of point A as functions of t:
x= 2 cos(3πt)
and y= 2 sin(3πt)
(b) Write a formula for the slope of the tangent line to the circle at the point A at time t seconds:
-cot (3πt)
(c) Express the x-coordinate of the right end of the rod at point B as a function of t: 2 cos(3πt) +41/1
(d) Express the slope of the rod…
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