00:01
Hi, here in this given problem this is the surface of the liquid and at the bottom of the pool which is filled with this liquid there is a flashlight.
00:16
This flashlight is making an angle theta with the vertical with respect to the vertical.
00:24
This is the light coming out of that flashlight making an angle theta which is missing, we have to find it.
00:30
So, the same will be the angle of incidence made by this ray with the normal drawn at the interface, that is angle of incidence.
00:42
Then the ray of light, flash light, goes into the air making an angle of refraction are index of refraction of the liquid that is given as 1 .36 angle of refraction in the air that is given as 31 degree angle theta angle of flashlight with respect to the vertical that will be equal to the angle of incidence so in order to find this angle theta using snail's law we can say the ratio of sign of angle of incidence to the sign of a refraction that is equal to the ratio of the refractive index of the second medium the second medium is air.
01:36
Means that is 1 to the index of refraction of the first medium which is the liquid nl.
01:45
So here it will become sine i.
01:48
It is rearranged to get sine i is equal to sine r and for r this is this is 31 degree divided by nl which is 1 .36 and it is calculated to be equal to 0 .3787.
02:02
So finally, this angle i will be given by sine inverse of 0 .3787 means this angle theta of the flashlight with respect to vertical which is equal to i that is calculated to be equal to 22 .3 degree.
02:23
One of the answer for this given problem here...