Assuming that you have a triangular prism, the ray of light will undergo refraction twice. The first time is the transition from air to flint glass on the entry face, and the second time is the transition from the flint glass to air from the exit face. With the available data, there are two possible solution since saying "20Âş from the normal" isn't enough information. Depending upon which side of the normal that 20 degrees is, the interior triangle will have the angles of 35, 90-r, and 55+r, or 35, 90+r, 55-r degrees where r is the angle from the normal after the 1st refraction. I will provide both possible solutions and you'll need to actually select the correct one based upon the actual geometry which I don't know because you didn't provide the figure or diagram that you were provided with.
The equation for refraction is:
(sin a1)/(sin a2) = n1/n2
where
a1,a2 = angles from the normal to the surface.
n1,n2 = index of refraction for the transmission mediums.
For this problem, we've been given an a1 of 20Âş and an n1 of 1.60. For n2, we will use air which at STP has an index of refraction of 1.00029. So
(sin a1)/(sin a2) = n1/n2
(sin 20)/(sin a2) = 1.00029/1.60
0.342020143/(sin a2) = 0.62518125
0.342020143 = 0.62518125(sin a2)
0.547073578 = sin a2
asin(0.547073578) = a2
33.16647891 = a2
So the angle from the normal INSIDE the prism is 33.2Âş. The resulting angle from the surface of the entry face will be either 90-33.2 or 90+33.2 depending upon the geometry. So the 2 possible triangles will be either 35Âş, 56.8Âş, 88.2Âş or 35Âş, 123.2Âş, 21.8Âş. with a resulting angle from the normal of either 1.8Âş or 68.2Âş. I can't tell you which one is correct since you didn't tell me which side of the normal the incoming ray came from. So let's calculate both possible exits.
1.8Âş
(sin a1)/(sin a2) = n1/n2
(sin 1.8)/(sin a2) = 1.6/1.00029
0.031410759/(sin a2) = 1.599536135
0.031410759= 1.599536135(sin a2)
0.019637418= sin(a2)
asin(0.019637418) = a2
1.125213477 = a2
68.2Âş
(sin a1)/(sin a2) = n1/n2
(sin 68.2)/(sin a2) = 1.6/1.00029
0.928485827/(sin a2) = 1.599536135
0.928485827 = 1.599536135(sin a2)
0.58047193 = sin a2
asin(0.58047193) = a2
35.48374252 = a2
So if the interior triangle is acute, the answer is 1.13Âş and if the interior triangle is obtuse, the answer is 35.48Âş
When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat
So here weight of the boat = Buoyancy force
let say boat is sunk by distance "h"
now we can say
now by above force balance equation we can write
so boat will sunk by total 5 mm distance
There are two equal forces of gravity between the Earth and the Moon.
One force pulls the Moon toward the Earth.
The other force pulls the Earth toward the Moon.
If only this gravity suddenly switched off, then the moon would
continue to orbit the Sun, very much as it does now.
If ALL gravity suddenly switched off, then . . .
-- the Moon would stop orbiting the Earth and would sail away, in
a straight line and at the speed it had when gravity disappeared;
-- the Earth would stop orbiting the Sun and would sail away, in
a straight line and at the speed it had when gravity disappeared;
-- all the gases surrounding the Earth ... which we call "air" ... would
start drifting away, and expanding into a giant cloud of gas, and stop
being an atmosphere;
-- the Sun would completely fall apart, expand into a giant cloud of gas,
and stop being a star.
Could you maybe show us the diagram
1) G
2) E
3) D
4) I
5) J
6) C
7) H
8) F
9) B
10) A
I think... i am not 100% sure....
