Cloud Shadow: CR4 Challenge (08/26/08)

Looks to me like additional information is required like time of day and latitude where the observation is taking place.

Evening shadows are longer than mid day shadows and further from the equator shadows are longer at high noon than equator shadows at the same time.

Regardless, I'd say B is always larger than A under any latitude or time of day situation.

Cloud shadows are not bigger than the clouds themselves. How can something so huge as the sun cast a shadow of a cloud bigger than the object itself. Has anyone ever seen the detailed mapping of moon's shadow when there is a solar eclipse coming? This is basically the same thing, according to the diagram. One object between the sun and earth creating a smaller shadow on the surface of the earth. Actually I guess according to the diagram this would all be fine, except that the sun is really bigger than the cloud.

Cloud shadows are not bigger than the clouds themselves.

If you're saying that a shadow can't be bigger than the object creating it, then you need to walk down a country road with a small child near sunset (with the sun behind you) and have the child describe what they see....

I agree with your first paragraph - yes, we need more information about the apparent position of the Sun to give even reasonable accuracy. On the other hand, we can add the position of the sun (relative to cloud and its edges) as a pair of angle parameters

But the lengthening you describe is due to the way that these common shadows are created; the object that casts these shadows is generally nearly vertical, and the ground on which the shadow appears is relatively horizontal. This effect relies on the surfaces not being parallel.
You can see an example of this with shadows cast on a wall by a vertical object; you will see that the vertical extent of the shadow is very similar to the vertical extent of the section of the object that created them.

While on the subject, I was wondering of the diagram was deliberately misleading, or merely constrained by the limitations of drawing. The sun is shown small and close, whereas in reality it is immense and distant. A small close light source will generally produce enlarged shadows as sketched in the challenge. A large but distant source will produce a blurred shadow, and if the surfaces are parallel a shadow defined by the middle of the "blurred" region will be only very marginally larger than the object. However, the drawing illustrates the inner edge of the blurred region - which will be appreciably smaller than the object (that is the extent of the darkest region of the shadow - which seems a reasonable position to take) .

At first light, this seemed like such a simple question to answer. But look at how the various disagreeing theories have clouded the issue! I say wait a few hours. By midnight, no one will care and all this will be forgotten...

What? Is there some sort of sunset clause in the CR4 terms of use that has expired, making serial puns no longer illegal?

With all thing being equal, assuming a perfect right angle between the sun and the cloud the shadow would be approx. 740.2m/2428.3ft or .46mi across... This would be a shadow approx. 14.8 times larger than the cloud itself......

Using a calculation based on similiar triangles I used the following data

Dia. Earth 12,900 Km

Dia. Sun 1.38x10^6 Km

Dist. Earth-Sun 149.6x10^6 Km.

Then (12,900+1.38x10^6)/( 2 * 149.6x10^6) = X/1500, X being the additional

Radius of the shadow. X = 6.983 m, so the shadow (b) is 13.983 meters larger

than the cloud (a). Actually looking at the edge of the sun behind a cloud you

see the left edge of the sun at the right edge of the cloud. The ray at the edge

of the shadow comes from the opposite edge of the sun.

The cloud is 1.5 kM from Earth's surface.

The sun is, on average, 149,597,890 km from the Earth's surface.

This means that for all practical purposes, we can assume that light rays from the sun are almost parallel when they reach the earth.

Hence, dimension B is only a minuscule amount bigger than A.

To a first approximation, it would be bigger by an amount equal to :-

149,597,890/(149,597,890-1.5) = 1.00000001002688

This means the shadow is approximately 0.5μm (Yes, that is half a micron) bigger than the 50m cloud.

The sun is half a degree wide so perhaps the shadow is smaller than the cloud?

You could be right here.

It does look from the diagram drawn that they are looking at the opposite extreme edges of the sun.

Would you like to do the calculations based on this assumption?

I do believe just what you have said. the size of the light source, its distance, and the proximity of cloud to earth makes me believe there's no noticeable size difference between the cloud and its shadow.

This is same problem as that of Total Solar Eclipse and the size of the shadow of the moon on the earth surface.

Considering average distance of Sun from Earth as 150,000,000 Kms (forgeting the acurarate figure you have stated, just for simplicity of the calculation, Sun diameter as 140000 Kms, and assuming the sun is directly overhead and the cloud is symmetrical around the visible Sun disc, we can calculate the size of the shadow of the cloud as 48.25 meters. (Assuming cloud as sharp edge object of 50 m diameter). Here we neglect the curvature of the earth, which will expand the effective shadow diameter on the earth surface.

The shadow will reach 42.857 Kms from the cloud, well below the surface of the earth. (Apex distance from the cloud)

We need not worry about the penumbra diameter, as it will be much bigger and will not be visible.

I do not think, I need to put the figure, as all results are from simple Trignometry. Should I?

I think you are, in principal, correct. However, the angle between the lines you show forming the antumbra would be 1 degree. So 50m / Tan 1^{o =} 2864m. That should place the crossing of lines at 1364 meters below the earth's surface, I think.

A scale model would make this all clear as mud. The thousand meter solar system walk makes the sun the size of a soccer ball, the earth the size of a peppercorn, and our cloud the size of a small dust speck. What to use to draw the pencil lines fine enough, even at a scale that puts the earth 25 paces away from the sun?

The Sun only subtends about 0.5-degree, which is the angle between the lines forming the antumbra.

If you take Oldr&wiser's more direct approach, an overhead sun would give a distance below the cloud of approximately (diameter_Cloud)*(distance_Sun)/(diameter_Sun) ≈50*149.6x10^9/1.38x10^9 ≈5420 metres. That would be 3920 metres below the Earth's surface.

But (I need to be careful here - I usually find I've made a doo-doo when supposedly correcting Randall) - I can't think where Randall's 6000-metres could have come from either.

The Sun only subtends about 0.5-degree, which is the angle between the lines forming the antumbra.

Yes, indeed. Foggy day here!

About 6 kilometres was just a guess. If the sun subtends ½ a degree, then the real picture looks a bit like this:-

Antumbra starts 25/Tan(¼º) below the cloud = 4230 metres below the earth.

d is 1500 * Tan(¼º) = 6.545 metres.

So the umbral shadow is 6.545 metres smaller than the cloud and the penumbral shadow is 6.545 metres larger than the cloud.

Your numbers are probably more correct as the ½º is probably just an approximation.

All assumes the cloud is directly overhead.

Thanks (I'm happy with approximations - in any case, the Sun-to-Earth distance is somewhat variable).
But wouldn't that be 6.545 metres on each side - i.e. ~ 13.1-metres larger than the cloud in total?

Correction for the angle of the sun from the surface normal looks to be reasonably straightforward. The extreme rays leave the cloud at θ+/-0.25^O from the surface normal, so the extreme distances of the umbra from the edge of the cloud are:
1500.tan(θ+0.25^O), and 1500.tan(θ-0.25^O) respectively (in the plane containing the central ray and the surface normal; and
+/-1500.tan(0.25^O)/cos(θ) in the vertical plane orthogonal to that.

So the size reduction is:
approximately* 300.tan(0.25^O)/(cos(θ))² along the line including the ray and the surface normal; and
ideally 300.tan(0.25^O)/cos(θ) along the orthogonal line.
*The error due to the approximation for the first case is less than 1% up to 87-degrees and 10% up to 89.2O

P.S. I assume your "cloud is overhead" is intended to mean the shadow is centred under the cloud - i.e. that the sun falls orthogonally on the ground?

For some reason, this submitted itself in the middle of composition. I didn't find out until I'd had my tea and returned and finished writing. It should have read:

Thanks (I'm happy with approximations - in any case, the Sun-to-Earth distance is somewhat variable).
But wouldn't that be 6.545 metres on each side - i.e. ~ 13.1-metres larger than the cloud in total?

Correction for the centre of the sun being an angle θ from the surface normal looks to be reasonably straightforward, at least along the principle directions. The extreme rays leave the cloud at θ+/-0.25^O from the surface normal, so the extreme distances of the umbra from the edge of the cloud are:
1500.tan(θ+0.25^O), and 1500.tan(θ-0.25^O) respectively (along a line that crosses the central ray and the surface normal); and
+/-1500.tan(0.25^O)/cos(θ) (orthogonal to that).

So the size reduction is:
1500.(tan(θ+0.25^O)-tan(θ-0.25^O)) along the line including the ray and the surface normal; and
300.tan(0.25^O)/cos(θ) along the orthogonal line.
*or, approximately, 300.tan(0.25^O)/(cos(θ))²; the error due to this approximation is less than 1% up to 87^O and 10% up to 89.2^O.

P.S.1. I assume your "cloud is overhead" is intended to mean the shadow is centred under the cloud - i.e. that the sun falls orthogonally on the ground?
P.S.2. The the Sun's image would be a perfect ellipse, so extension to other angles should be straightforward (albeit too tedious for me today).

But wouldn't that be 6.545 metres on each side - i.e. ~ 13.1-metres larger than the cloud in total? Yes

P.S. I assume your "cloud is overhead" is intended to mean the shadow is centred under the cloud - i.e. that the sun falls orthogonally on the ground? Yes again: should have read "sun is directly overhead"

Must learn to be more careful.

Yes, I agree the relative distance of the sun from the earth does make the light rays nearly parallel, but the sun is not a point source of light. The sun is very wide compared to the earth and especially the cloud. Toss in the diffusing effects of the earth's atmosphere and I would say the shadow is slightly smaller at best.

The diagram clearly shows the Sun with significant diameter and the rays depict the region where none of the sun's light arrives; so I think we can assume the question is directed towards the size of the umbra.

As the image would suggest, it is high noon and the cloud is directly between the sun and the shadow. This assumption must be made because we are not defining the orientation "B" has to the earth's geometry. The sun is many times larger than the cloud. Rays generated from the edges of the sun would illuminate the earth directly underneath the cloud resulting in the shadow being smaller than the cloud itself. The sun's rays are not actually parallel but radiate from all points of the sun's surface in all directions.

Ted Rauhut

so the umbra should not reach earth

The cloud is about four times as big as the sun (in an angular sense), so, why won't it completely blot it out? Otherwise, good answer.

Dear friends,

1. At noon, the size of the shadow will be equal to the size of the cloud.(Considering the distance of sun and straight beam of light)

2. At other times, the the size of shadow is inversally proportional to the angle made by the sun with the azimath/ horizon of the earth(viewer) and directly proportional to its altitude (though negligible).

3. Very small sized cloud at very high altitude will not cast shadow due to bending of light rays along the object to regain their normal path.

Haaving chased the shadow of balloons and airplanes, from anecdotal evidence, I think the shadow is smaller than the cloud,

I agree with Uncle Red - the difference in size will depend on the angle at which light strikes the cloud and the earth. It will also depend on the effective thickness of the cloud. As the height of the cloud is given as a single number, I think for the purposes of this problem we must assume that the cloud is thin and parallel to the ground.

On the basis that the correction given by Maths_Physics_Maniac is even smaller than corrections due to the Earth's surface curvature and to atmospheric lensing, which in turn are smaller than the variation in sizes due to the eccentricity of the Earth's orbit, I think that we should ignore all four of these effects.

If the sun's rays arrive perpendicular to the cloud and ground surfaces, I believe that Oldr&wser and Blink have the basics of the answer; however, Oldr&wser's answer is incomplete, and (as stated by Randall) Blink appears to have mistyped the terminology in his answer.
So, assuming that the ground and the cloud are parallel, etc, and the sun is at an angle of θ to the surface normals:
the darkest part of the shadow (the umbra) will be about 1500*SunDiameter/SunDistance/cos(θ) ≈ 13.8/cos(θ) metres smaller than the cloud (or somewhat smaller than 37-metres actual size); and
the region where the sun is at least partially obscured by the sun (the penumbra plus the umbra) will be 13.8/cos(θ)-metres larger than the cloud (or somewhat larger than 67-metres actual size).

N.B. Ignoring cloud thickness, the sun would need to be on a slant of at least 74-degrees from the vertical before the cloud would be completely enclosed by the sun (at which point there would be an antumbra). (Alternatively, if the sun was overhead, the cloud would need to be at a height of > 5400-m)

N.B.2 To illustrate the practical effects - in New York at the equinox and midday these differences in size would be about 18.3 metres. Without further definition, I would take "shadow" to mean the region from which all of the sun is obscured by the cloud, so I would give my short answer as "possibly about 18.3-metres smaller than the cloud".

Apologies, I didn't fully interpret the question, and off-angle effects result in a larger size reduction than given here. The diagram in the challenge clearly shows the extreme rays from the sun defining the shadow, so the size required is the umbra - that means that the shadow shrinks. It also shows the sun is offset in a plane including the direction in which the size of the shadow is to be measured. So the solution should read:

Shadow size reduction ≈
1500.((tan(θ+0.5^O)-(tan(θ+0.5^O))/cos(Φ),
where θ is the angular offset of the central rays of the sun in the plane of the challenge drawing, and Φ is the offset of this plane from the ground/cloud perpendicular. (N.B. The 0.5^O is approximate).

It will depend on latitude and the time of day or whether or not the sun is directly above the cloud. If the sun is directly above the cloud, at the equator at high noon, it will be .93205/93,000,000 or 1.202 x 10-8 times larger. For all practical purposes the same.

I've read the posts of others and I'm impressed with the quality of thought. My suspicion is that the devious minds of the challenger's have selected the height of the cloud to be exactly correct such that the cloud's shadow is exactly the same size as the cloud given the reasoning and geometry presented by Randall in #13.

However, this would only be the correct answer when the sun is directly overhead and the cloud is at the equator. Which represents a cloud shadow minim. Then there is the boundary condition issue. What constitutes a shadow? The size of the sun means there is a gradient of shadow from penumbra to umbra. If only the umbra - the darkest shadow counts then the cloud shadow would be smaller than the cloud even at noon on the equator.

Seems to me that there are significant gray areas in this challenge question. (Pun intended)

Silly me,

Mr. Gee

The mass of water droplets (cloud) will act as a lens so that light passing by cloud's edge at normal angle or obliquely will be directed inward (more or less) towards the central vertical "axis" extending earthward from the cloud...thus lighting up (and thereby "obscuring") the (otherwise) periphery of the cloud shadow at ground level...that is, diminishing shadow expanse to no greater than that of the cloud itself.

just thinking of 50 meter cloud at a height of 1500 meters!

I cannot figure out why people see the sun as a point source of light, since everyone must have seen just an edge of the sun and be blinded by it. Now you can all do trig. so we all can calculate the width of the umbra and penumbra. The questions are never very specific in order to get us in a wide discussion. I like it.

Now... what is a shadow. It is a reduction of light due to an object. So the "shadow" size is the penumbra shadow.

If one takes another path and says the "shadow" is the umbra shadow. I think (assuming no light will penetrate the cloud) you are thinking of the shadow as no light. Well, these people have another problem. On that assumption there is something called diffraction. When you assume the cloud is an exact circle there will be various circles without light almost 0 in width. With a form like a cloud I state that there will be no area without light at all.

the shadow is ≈ 2 mm smaller than the cloud.

***

angle = arctan(radius of sun/ distance of sun to earth) = .00004°

shadow (m) = 2*1500*tan(.00004)

I am assuming the center of the cloud is aligned with the center of the sun and earth.

A 50 m cloud is very small compared to the sun and the earth so the curvature of the earth is negligible. The length of the cloud doesn't matter, so this would be correct for any position on the earth, provided the sun is still shining.

I screwed that up.

the answer is ≈ 14 m

the angle isn't .00004 it's .27° -- big difference.

I voted you a conditional GA, because you imply the shadow will be smaller than the cloud width. But by how much?

This depends on where you are, the time of day and year and which direction you measure the shadow. That is why an equation is needed.

Here are a couple of examples of the reduction* in size at some locations and times:

Equator, mid-day, solstice (i.e. sun overhead), flat ground: about 13.8-metres.
New York, mid-day, solstice, flat ground: 18.6 metres if you are measuring East-West, 25.1-metres if you are measuring North-South.

If the sun is falling at more than about 58^O 20' degrees from the vertical, and the cloud is a 50-metre circle, the shadow will just vanish.

*The values are based on the average distance and diameter of the sun (rather than the 0.5-degree angle).

And that's what I was getting at - there apparently are no conditions or circumstances under which the shadow diameter will appear to exceed the cloud diameter, correct? The lessened apparent diameter will be variable, but always lessened.

BTW, this is what I thought (but did not express) from the beginning, based on observation. I have often watched the shadow of an aircraft I was a passenger in against various backgrounds, and it always appeared to be markedly smaller than the plane itself. I did not have the math available to explain why, so I didn't intrude on the discussion.

I assume we are talking about shadows cast by the sun. Even then it depends on the position, size, and orientation of the object casting the shadow; physiology is involved just as much as physics.

The shadows of objects that are parallel to the ground and whose vertical extent is very small will always appear smaller than the objects themselves (I'll go more into the reason for this in the next paragraph). However, if the object is at an angle to the ground or has significant vertical extent it can appear enlarged - an example would be the shadow of an aeroplane immediately after take-off if the sun is behind it.

The reason the first class shadows appear reduced in extent is possibly as much due to human physiology as the properties of light. For now, I'll restrict consideration to the case where there is a region of shadow from which the sun cannot be seen at all (this is called the umbra). Then there is a region of shadow from which part of the sun can be seen - the penumbra. The umbra is the part that is smaller than the original object. The outer edge of the penumbra is larger by the same amount as the umbra is smaller (measured linearly). Within the penumbra, the light level increases monotonically from the outer edge of the umbra to the outer edge of the penumbra - there are no sudden changes.
However, we still see a well-defined edge to the shadow - at the boundary between the penumbra and the umbra. This is because the visual system can detect edges by first adjusting to the local intensity and then looking for changes in slope - which means that the sharpest perceived change is at the edge of the darkest (and smallest) region.

P.S.

I assumed that you were writing about the aeroplane appearing thinner rather than it appearing significantly shorter, because (typically) the length is a large multiple of the width, so the shadow would become quite indistinct well before the reduction in apparent length becomes significant (even a jumbo is more than ten times longer than it is wide).

If it appears shorter, there can be a number of reasons, for example that the thinner parts no longer have an umbral region. But I think it at least as likely that the shadow appears small because the distance to the shadow is greater than is interpreted by your visual system. (Note just how small a plane at typical cruising height appears from the ground - although of course that will not cast a proper shadow at ground level)

Fyz

What is your answer?

My answer is , no matter where the cloud is, the shade is alway equal to itself.

the earth plane can be seen 25km. 5om is very samller than this size.

the sun is more larger than earth, the ray from it can be see parallel ray. so they are always the same.

why do you list so many math formula?

It is not the size of the Earth or the cloud that are important for this, but the size of the Sun.

The Sun may be very far away from the Earth, but the Sun is so large that is still subtends an angle of about 0.5^O. So the shadow will be blurred. The size of the blurred region will depend on the distance to the ground along the sun's rays, and the angle at which the rays hit the ground. It will not depend on the size of the cloud.

The usual definition of the size of such a shadow is the size of its darkest region. As the drawing that is used to clarify the challenge also corresponds to the ray coming from opposite edges of the sun, it seems we must take this definition. Using this definition for shadow, we find it must be at least 13-metres smaller than the cloud. That is not trivial compared to a 50-metre diameter cloud - and it is the amallest possible reduction in size, and only happens if you measure the reduction when the Sun is vertically overhead.

If the Sun is not overhead, the distance along the light path from cloud to ground will be increased, so the blurring of the shadow will increase - resulting in a further decrease in the size of the darkest region. Similarly if the rays hit the ground at an angle, the spread on the ground will be larger than the spread perpendicular to the rays.
These are the effects that I modelled. You can judge the significance from the fact that if the sun is falling at more than 58^O 20' from vertical a 50-metre diameter cloud would not cast a "proper" shadow at all. So no shadow in early morning or late afternon anywhere in the world; and even places like Lisbon, Bejing and Algiers would not see a proper shadow at any time of day near mid-winter. Even in summmer, Helsinki will only get a shadow very near midday.
So, you need something of the kind to get any idea that this might happen - and equations are the easiest and most precise way of working.

Why not take account of shape of the earth?

suppose if the cloud was projected on a surface, say sphere surface, what is the area of the cloud shade?

so we have to suppose the the project area is plane. As it was. in the 25km arrange, we can say, its plane. then we take account of sun size and its distance. thats all why we get in such result.

very simeple, just do a experience under sunlihgt on a suny day.can prove above theory.

Perhaps we would do better to scale the experiment down a bit so we knew what we were measuring. 25-cm cardboard projected on a stick from a second-floor window to give 7.5-metres, perhaps?

you do make a kidding. sun is not as a lamp as close to cardboard.

look at your #43. u have given out the solution, but you make a simple question complex.

don't consider semi shadow as do as a finite surface light source do as well as dot light source.

the sunlight is parallel light source and cloud is parallel to ground as well.

don't consider scatter light. it acts little action.

this can be proved by standing a stick on the ground. you can only see one shadow, not a area. or two or threes or more line shadows. this shadow shifts according to sun moving.

Either we are talking at cross purposes, or you are not working to follow what I and others write; neither are you looking very carefully at the shadow of your stick. The shadow of the bottom of the stick will be very sharp, but you will be able to see a region of semi-shadow outside the central region in the shadows formed from the higher points on the stick - this should be obvious once this approaches about 1-metre above the ground (you will usually see blurring rather than multiple shadows, because the sun is a solid disc and not separated sources).

In case it helps: stars other than the sun really are so far away that they can be approximated as point (or dot) light sources, and that is how they appear when you look at them. But, as you should have seen from every picture of the sun, or even observing the sun at sunset when it is safe to look for a short time, the sun occupies quite a width in your field of view.

It escapes me why you feel that argument is better than actually trying the experiment...

Regarding "overcomplexity" of post #43: the challenge effectively asks "how much size increase?" As the correct answer to the challenge is "there will be a size reduction", you are right that a literal interpretation doesn't really need this much detail. On the other hand, if we want to know how much the size of the shadow reduces, this is about the minimum we can use. (Strictly speaking, it was a a correction to previous comments - because these would only be correct if the sun was vertically overhead)

I have either observed carely the shadow of a stick in sunlight, nor read clearly above threads. what I hope to say is sun is not only disc shape but large enough, we well as remote and thus, give out a parallel light ray.

this ray and gound compose of an affine coordinates. not cardesian coordinates we see ordinarily

suppose the sunlight pass through a wondow, the situation is another thing, which is similar with your words.

This will unfortunately come over as condescending. If we were on a single site, we could very easily work together and overcome any language difficulties and agree the the relevant features.

Rays from any point on sun are effectively parallel. The angle between rays originating from opposite edges of the sun's disc is approximately 0.5^O - not at all parallel.

Regarding your experiment with the stick - you clearly need to look in more detail. Perhaps the least ambiguous version of this experiment:
Get two similar cylindrical sticks about 1-metre long and of uniform width, and place one upright in the ground. On a clear day, and when the sun is at or below 45-degrees to the vertical, place the second stick upright in the ground centred in the shadow of the first, and about 2-cm inside the far end of the shadow cast by the first stick. Observe the illumination of both edges of the second stick. Observe also the change in the characteristics of the shadows - compare the shadow of the first stick just before it reaches the second stick with the shadow of the second stick just after the end of the shadow of the first stick.
Please do not reply until you have actually done this.

The size of the blurred region will depend on the distance to the ground along the sun's rays, and the angle at which the rays hit the ground. It will not depend on the size of the cloud.

'Size' : It's annular and a function of cloud radius ?

I could have said "width perpendicular to the local cloud edge", but you already knew that from the context.

Again the official solution lacks the depth and flavor of the discussion. I think we should often consider the challenge question as the rock used for rock soup. It starts the process going. It is the substance, energy, and knowledge brought to the blog by the individual contributors that provide most of the substance. Admittedly, without a good starting focal point we wouldn't start this process and coming up with good challenges weekly must be a difficult task. I surely hope whoever is tasked with that job keeps doing it. I sure like to participate in this learning activity.

Again the official solution lacks the depth and flavor of the discussion.

You should have gone into in politics!

The official answer would suggest that a 10 m cloud at 2500 m would cast a 10 meter shadow. In fact, (by blocking about 1/2 the diameter and 1/4 the area of the sun) it would cast a large penumbral shadow and no umbral shadow at all.

So how does one measure the height of a tall building with a barometer?

"So how does one measure the height of a tall building with a barometer?"

I would take the barometer to the building custodian and tell him "I will give you this fine barometer if you will trade me a copy of the blueprints for your building showing the exact elevation of the structure." Done?

So, have you guys reached your verdict yet or just slowly giving up on this issue?

A lot of you mentioned that 'for all practical purposes' the size of the shade of the cloud should be the same, providing it gets targeted buy the Sun near directly from above as it looks in the Question.

1. Some of you stated that due to the vast distance of the Sun to Earth the light-rays arrive more or less parallel to the earth's surface
2. If the cloud is dark and the light doesn't get through, than it should be regarded as any normal object that casts an assumed perfect shadow.

Since light is highly directional (doesn't bend) therefore, with the aforementioned factors 'for all practical reasoning' it can be said, the shadow the cloud is casting onto the surface of the earth is more or less the same - without the shadow of doubt.

Those who do not believe it can or should go and check it.

How you gonna do it, is up to you.

The Sun may be a vast distance away, but it is also a vast size - so you can think of it as a large number of light sources spread across an angle of about 0.5^O. (Be honest - does the sun look as tiny as a star)

Now think about standing on the ground looking at the sky:
If you can see part of the Sun obscured by the edge of the cloud, you are in partial shadow.
If you only see the Sun as a glow through the cloud, you are in "full" shadow.
If you can see all of the Sun unobscured, you are not in the cloud's shadow at all.
Obviously, the region of full shadow is smaller than the region that is in some sort of shadow; nearly as obvious is that one of these is wider than the cloud and that the other is narrower than the cloud.

So, to answer the challenge, you need to determine the challenge's definition of shadow. The rays drawn in the diagram of the challenge come to the ground at the edge of the region of full shadow, so that must be what the challenge means.

I.e. the shadow as defined in the challenge is full shadow, and I say it is smaller than the cloud.

If you do not believe this, you could have the humility to check it for yourself. Demonstrating the principle should be quite straightforward if you use a long strip of cardboard on a bright day. A 4-cm wide strip of card held 2-metres above the ground should be adequate, provided the sun is not too low in the sky. But be careful that the card is parallel to the ground or you could confuse the issue. (If the sun is not vertically overhead, I would recommend pointing the long direction in the general direction 'towards' the sun in the first instance - you can go on to see what happens with other directions as a separate exercise).
It will be even more illuminating if you come back and tell us your results. (As an apparent disbeliever in penumbration, your report of its existence will carry far more weight than ever mine could)

You don't seem to keep the ratio between the size of the card and the distance as wrt the size of the cloud and its proximity to earth.

Like most people said it before, if anything the shadow is lightly smaller than the cloud but for all practical purposes it is negligible.

Why it should be smaller instead of larger? PROVIDING WE ASSUME PARALLEL LIGHT RAYS, light scatters including LASER light that is the most coherent light. Remember, nothing is perfect but sometimes can be assumed perfect for practicality as long as the dimension allows it!

Where did you get the 0.5 degree angle of light incidence from?

If you tell me all these nitty-gritty bits than I am all ears, no kidding.

I think I can explain the half-degree angle. That is the angle subtended by the apparent width of the sun. It isn't a great deal, but it does elevate the solar light beyond that of a point source. Light from the observed left side of the solar disc travels a different path than light from the observed right side, and both sides shine to the left and right as well as straight "down" toward Earth. So you get light shining from the left of the sun at both the left and right sides of the cloud, not just the center of it. Also from the right side at both the left and right of the cloud.

I hope this sheds a little light on the question you asked...

In that case the shadow considerably smaller than the cloud itself.

Like most people said it before, if anything the shadow is lightly smaller than the cloud but for all practical purposes it is negligible.

In practical terms it is not negligible. Take a disc of cardboard of 5 cm to represent the cloud, and hold it 1.5 meters above ground level on a very clear day, at a time close to solar noon. Place a piece of graph paper on the ground so you can see the size of the shadow easily. Ideally, prop up the graph paper so that it is normal to the sun, and keep the 5 cm disc normal to the sun too. When the disc is within 10 - 20 cm of the paper, the shadow will appear very distinct. as you move tht disc to 1.5 meters, the umbral part of the shadow will shrink, and the penumbral part (which is visible on a clear day against bright paper) will get larger. The umbral portion will be 13 mm smaller that the disc.

The official answer is simply wrong, because it ignores the fact that the sun is not a point source. Just look at any kindergartner's drawing of sun, clouds and trees.

I give up. I should have started out here before all arguement.

Given and taken minor errors, if we assume the incoming rays are parallel to both cloud and earth's surface then A~B i.e. the same.

the answer is published on 2rd sept. ?

why dont we see it even on 9th sept. ? and I dont see it until today.

what did on earth it published out?

Look more carefully - a very short (and very wrong) 'official answer' was posted some time ago.