I am trying to help my step son who missed half of physics class due to a funeral.
A laser beam is being shot through two different materials that have been stuck together. if theta 1 = 65.5 and theta 2 = 49.8 what is the index of refraction? (I got 1.2 which is correct) but cant get part 2
PART 2~ If the index of refraction of n3 is 3.6, what is the final angle of transmitting light theta 3?
THANKS IN ADVANCE FOR ANY HELP!
Chad on behalf of Zach ;)
To solve part 2 of the question, we can use Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the materials involved. Snell's Law is given by:
n1*sin(theta1) = n2*sin(theta2)
Where n1 and n2 are the indices of refraction of the respective materials, and theta1 and theta2 are the angles of incidence and refraction.
In part 2 of the question, we have n1 (index of refraction of the first material) as the calculated value of 1.2, theta1 as 65.5 degrees, n2 (index of refraction of the second material) as 3.6, and theta2 as 49.8 degrees.
We need to find theta3, the angle of transmission when the light passes from the second material to the third material, which has an index of refraction of n3 = 3.6.
To find theta3, we can rearrange Snell's Law as:
sin(theta2) = (n1/n2) * sin(theta1)
By substituting the given values, we can calculate sin(theta2):
sin(theta2) = (1.2/3.6) * sin(65.5)
Now, to find theta3, we again use Snell's Law:
n2 * sin(theta2) = n3 * sin(theta3)
Rearranging this equation to solve for sin(theta3), we get:
sin(theta3) = (n2/n3) * sin(theta2)
By substituting the values, we can calculate sin(theta3):
sin(theta3) = (3.6/3.6) * sin(theta2)
Finally, we can find the value of theta3 by taking the inverse sine of sin(theta3):
theta3 = arcsin(sin(theta3))
Calculating this value will give you the final angle of transmitting light, theta3.
Please note that it would be helpful to have a scientific calculator or a calculator with trigonometric functions to perform the necessary calculations accurately.