A ray of yellow light ( f8= 5.09 × 1014 Hz) travels at a speed of 2.04×10 meters per second in
2 answers:
I will assume that question in which medium the light of that frequency travels at that speed.
To find this answer we will have to calculate the refractive index of that medium.
Refractive index of a material is a number that tells us how light propagates through a medium. It tells us how much ray of light would bend entering the medium from a vacuum. It is defined as:
c is the speed of light in the vacuum and v is the speed of light in the medium. Now we can calculate refractive index:
We can search for a material that has this refractive index.
I found that Pyrex, glycerine and vegetable oil have a refractive index of 1.47.
To find this answer we will have to calculate the refractive index of that medium.
Refractive index of a material is a number that tells us how light propagates through a medium. It tells us how much ray of light would bend entering the medium from a vacuum. It is defined as:
c is the speed of light in the vacuum and v is the speed of light in the medium. Now we can calculate refractive index:
We can search for a material that has this refractive index.
I found that Pyrex, glycerine and vegetable oil have a refractive index of 1.47.
The complete question in the attached figure
Let
c ------------------- > is the speed of light
v ------------------- > is the speed in medium
n ------------------- > is the refractive index of medium
we know that
c/v = n
n = (3 x 10^8)/(2.04 x 10^8) = 1.47
1.47 is the refractive index of glycerol.
therefore
the answer is (4) glycerol
Let
c ------------------- > is the speed of light
v ------------------- > is the speed in medium
n ------------------- > is the refractive index of medium
we know that
c/v = n
n = (3 x 10^8)/(2.04 x 10^8) = 1.47
1.47 is the refractive index of glycerol.
therefore
the answer is (4) glycerol
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Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:
So, the new pressure is 2 atm.