Question

A laser beam in air is incident on a liquid at an angle of 35.0 degrees with respect to the normal. The laser beam's angle in the liquid is 26.0 degrees. What is the refractive index of the liquid?

A) 1.09
B) 1.35
C) 1.28
D) 1.50

Answer 1

The refractive index of the liquid is 1.35.

Step-by-step explanation:

To find the refractive index of the liquid, we can use Snell's law:

n1sin(θ1) = n2sin(θ2)

Where:

• n1 is the refractive index of air (1.00)
• θ1 is the angle of incidence (35.0 degrees)
• n2 is the refractive index of the liquid (unknown)
• θ2 is the angle of refraction (26.0 degrees)

Plugging in the values:

1.00sin(35.0 degrees) = n2sin(26.0 degrees)

Solving for n2:

n2 = (1.00sin(35.0 degrees)) / (sin(26.0 degrees))

n2 = 1.35

Therefore, the refractive index of the liquid is 1.35.