A laser beam in air is incident on a liquid at an angle of 40.0^\circ with respect to the normal. The laser beam's angle in the liquid is 25.0^\circ.What is the liquid's index of refraction?
To find the liquid's index of refraction, we can use Snell's Law, which relates the angles and indices of refraction of light when it travels between different media.
Snell's Law states:
n1 * sin(theta1) = n2 * sin(theta2)
where:
- n1 and n2 are the indices of refraction of the initial and final media, respectively
- theta1 and theta2 are the angles of incidence and refraction, respectively, with respect to the normal.
In this case, the laser beam is incident on the liquid at an angle of 40.0° with respect to the normal, and the angle of refraction in the liquid is 25.0°.
We need to calculate the index of refraction of the liquid (n2).
Using Snell's Law, we can rearrange the equation to solve for n2:
n2 = (n1 * sin(theta1)) / sin(theta2)
Plugging in the given values:
n1 = index of refraction of air (which is approximately 1.00)
theta1 = angle of incidence in air = 40.0°
theta2 = angle of refraction in the liquid = 25.0°
n2 = (1.00 * sin(40.0°)) / sin(25.0°)
Now we can calculate n2 using a scientific calculator or an online calculator:
n2 ≈ 1.36
Therefore, the liquid's index of refraction is approximately 1.36.
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