What Surprises Await the Observer in These Physics Scenarios?

[Andy_ToK]

Homework Statement

Q1. A little girl is holding a helium-filled balloon on a string while
riding in a closed elevator going down a very tall building at
constant speed. The elevator shaft is maintained under vacuum.
Suddenly the elevator cable snaps, sending the elevator
into free fall. In her surprise the girl let's go of the string. She
is even more surprised to see
(a) the balloon rising.
(b) the balloon floating downward.
(c) the balloon remaining stationary.
(d) the balloon bouncing slowly between the floor and the
ceiling.

Homework Equations

Newton's 2rd and 3rd laws

The Attempt at a Solution

The answer is C which confuses me.
That means the balloon itself is accelerating in a stationary frame, then there must be some force acting on the balloon. Would it be the air in the elevator that is exerting the constant force on the balloon?

Homework Statement

Question 10
At the shallow end of a swimming pool, the water is 70.0
cm deep. The diameter of the cone of light emerging from
the water into the air above, emitted by a light source 10.0
cm in diameter at the bottom of the pool and measured by an
observer on the edge of the pool 2.50 m away, is’
(a) 1.60 m. (b) 1.70 m.
(c) 1.75 m. (d) 1.80 m.

Homework Equations

n1sin(a1)=n2sin(a2)

The Attempt at a Solution

Sorry, but i don't get what it means by "light cone" in the question


Thanks in advance!

 

[chemisttree]

1. In free fall, the force of gravity goes to zero within the reference frame of the elevator itself. All objects, including the air, are accelerating at a rate equal to the acceleration of gravity. The balloon has no weight... it does have mass. Since it is weightless there are no bouyancy forces acting on it.

2. I believe you are being asked about the angle at which all of the light from the light source is being internally reflected by the water. The light will strike the water/air interface and be refracted at an angle coincident with the surface of the water.

 

[Andy_ToK]

Thanks. But can you elaborate on the second question?
I thought I'm supposed to calculate the distance between the two light rays coming from two ends on the circular light source(the closest point to the edge and the farthest). But it seems that if so, the diameter of the light cone would be infinite at the observer.

 

[chemisttree]

I think that this site will explain it much more effectively than I can.

http://www.ps.missouri.edu/rickspage/refract/refraction.html

 

[Doc Al]

cone of light

Hint: This problem involves the critical angle. Imagine light emitted in all directions--only a portion will make it into the air, the rest will be reflected back. Find the diameter of the cone of light emerging at the water's surface.

 

[Andy_ToK]

Hint: This problem involves the critical angle. Imagine light emitted in all directions--only a portion will make it into the air, the rest will be reflected back. Find the diameter of the cone of light emerging at the water's surface.

Thanks, that's exactly what i was doing but i got the infinite diameter.
the angle of total reflection is 48, so as long as the incident angle is smaller than that, the light rays will come out of the water. Obviously, there'll be light rays traveling just above the water surface, so we only need to calculate the other boundary of the light cone and that's one i got trouble into. when the incident angle is close to 0, the angle of refraction is also close to 0, which means the light ray come out of the water surface is almost perpendicular to the surface, resulting in a light cone of infinite diameter.
I guess I misunderstood something. Please point it out. Thanks!

 

[Doc Al]

I see what you're thinking. But realize that all the light that makes it to the air looks as if it came from within a circle on the water surface. That circle is quite finite--find its diameter using what you know about critical angles. (Although I see what you mean about the "cone" going off at about 90 degrees.)

 

[Doc Al]

Check this out: http://books.google.com/books?id=hz...XE&sig=2eHsRZpcdqlkJJr7qQJ9axzWJo8#PPA151,M1"

 

[Andy_ToK]

oh, ya, i guess i misunderstood the question. the question is asking the diameter of the light cone "measured by the observer" not the "actual" one. oops,
but still no clue how to calculate it. any other formulas required?

 

[Doc Al]

any other formulas required?

Nope. All you need is the critical angle and some trig. Draw yourself a picture!

 

[Andy_ToK]

But wouldn't the height of the observer matter?
Thanks for the input.
I tried to draw the picture. I assume you meant the picture should be similar to the one on the link you gave me? To be honest, I really don't know how to draw it or I can't get the answer from the pictures I drew.

EDIT: i know "how to get the answer" now, by calculating the diameter of the "light disk" on the surface of water. But I'm not sure if that's right way to do it nor I understand why I should do it in that way. Will the observer actually saw a cone of light? or just a "shining disk" ?
The distance between the observer and the light source doesn't affect the result, but why?
Thanks!

 

[Doc Al]

Will the observer actually saw a cone of light? or just a "shining disk" ?

A very good question! If the air were dust-free, I'd say that the observer would not see a cone of light. (For the same reason that you don't see a beam of light emitted from a flashlight--unless there's dust in the air to reflect the light.) All any observer actually sees is the light that makes it to his eyes.