It's c/f, in which c is the speed of light (300000 m/s)
Answer: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4 - Chemical Equation Balancer
Equation is already balanced.
Explanation: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4
Your teacher is right. The moon can be seen early in the morning sometimes and late at night. Different phases are only visible on certain days as one day might be full quarter, the next full moon, the next first quarter, etc.
Answer:
(a) ![n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}](https://tex.z-dn.net/?f=n_%7B2%7D%20%3D%20%5Cfrac%7Bn_%7B1%7Dsin%5Ctheta_%7B1%7D%7D%7Bsin%5Ctheta_%7B2%7D%7D)
(b) ![n_{2} = 1.349](https://tex.z-dn.net/?f=n_%7B2%7D%20%3D%201.349)
(c)
(d)
Solution:
As per the question:
Refractive index of medium 1, ![n_{1} = 1.47](https://tex.z-dn.net/?f=n_%7B1%7D%20%3D%201.47)
Angle of refraction for medium 1, ![\theta_{1} = 59^{\circ}](https://tex.z-dn.net/?f=%5Ctheta_%7B1%7D%20%3D%2059%5E%7B%5Ccirc%7D)
Angle of refraction for medium 2, ![\theta_{1} = 69^{\circ}](https://tex.z-dn.net/?f=%5Ctheta_%7B1%7D%20%3D%2069%5E%7B%5Ccirc%7D)
Now,
(a) The expression for the refractive index of medium 2 is given by using Snell's law:
![n_{1}sin\theta_{1} = n_{2}sin\theta_{2}](https://tex.z-dn.net/?f=n_%7B1%7Dsin%5Ctheta_%7B1%7D%20%3D%20n_%7B2%7Dsin%5Ctheta_%7B2%7D)
where
= Refractive Index of medium 2
Now,
![n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}](https://tex.z-dn.net/?f=n_%7B2%7D%20%3D%20%5Cfrac%7Bn_%7B1%7Dsin%5Ctheta_%7B1%7D%7D%7Bsin%5Ctheta_%7B2%7D%7D)
(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:
![n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}](https://tex.z-dn.net/?f=n_%7B2%7D%20%3D%20%5Cfrac%7B1.47%5Ctimes%20sin59%5E%7B%5Ccirc%7D%7D%7Bsin69%5E%7B%5Ccirc%7D%7D)
![n_{2} = 1.349](https://tex.z-dn.net/?f=n_%7B2%7D%20%3D%201.349)
(c) To calculate the velocity of light in medium 1:
We know that:
Thus for medium 1
(d) To calculate the velocity of light in medium 2:
For medium 2: