A laser beam passes from air into a 25% glucose solution at an incident angle of 37 ∘ . In what direction does light travel in the glucose solution? Assume the index of refraction of air is n = 1.
From the reference book: the index of refraction
of 25% glucose solution is n₁=1.37
sin i/sinr=n₁/n=n₁
sinr= sin i/n₁= sin37°/1.37=0.439
r=sin⁻¹0.439=26°
Ah, the beam of light is taking a little detour into the glucose solution! Well, when light passes through different mediums, it likes to change its direction a bit. In this case, since the incident angle is 37 degrees, the light will bend towards the normal line (an imaginary line perpendicular to the surface) as it enters the glucose solution. So, it will be traveling towards the normal. But don't worry, it's just light trying to make its way through the sweet world of glucose!
To determine the direction in which light travels in the glucose solution, we need to compare the indices of refraction of air and the glucose solution.
The index of refraction (n) is a measure of how much light is bent, or refracted, when it passes through a medium. According to Snell's law, the angle of refraction (θ2) can be determined using the following formula:
n1 * sin(θ1) = n2 * sin(θ2)
Where:
n1 is the index of refraction of the incident medium (air),
θ1 is the incident angle,
n2 is the index of refraction of the second medium (glucose solution), and
θ2 is the angle of refraction we want to find.
Given:
n1 (index of refraction of air) = 1
θ1 (incident angle) = 37°
We need to find n2 (index of refraction of glucose solution) and θ2 (angle of refraction).
The incident angle and refracted angle are measured with respect to the normal, which is perpendicular to the surface of the medium. In this case, the normal will be within the plane perpendicular to the glucose solution.
Since we are looking for the direction of light in the glucose solution, we are interested in the refracted angle θ2.
To find θ2, we need to find n2. The equation can be rearranged as follows:
sin(θ2) = (n1 * sin(θ1)) / n2
Now, we can substitute the known values into the equation:
sin(θ2) = (1 * sin(37°)) / n2
To find n2, we need to know the index of refraction of the glucose solution. Since the percentage glucose solution is given, we can assume the index of refraction for a 25% glucose solution is approximately 1.38.
Substituting n2 = 1.38 into the equation:
sin(θ2) = (1 * sin(37°)) / 1.38
Using the inverse sine function to find θ2:
θ2 = sin^(-1)((1 * sin(37°)) / 1.38)
Calculating this value will give us the angle of refraction in the glucose solution, which indicates the direction of the light in the glucose solution.
To determine the direction of light in the glucose solution, we need to consider the principles of refraction and Snell's law. Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media. Mathematically, this can be written as:
n1*sin(θ1) = n2*sin(θ2)
Where:
- n1 is the refractive index of the first medium (air),
- θ1 is the angle of incidence,
- n2 is the refractive index of the second medium (glucose solution),
- θ2 is the angle of refraction.
In this case, the incident angle (θ1) is given as 37°, and the refractive index of air (n1) is approximately 1.
To find the refractive index of the glucose solution (n2), we need to know the speed of light in glucose. Since it's not provided, we can make an estimation using the given concentration (25%).
First, we need to find the refractive index of pure water (assuming the glucose solution has similar properties to water) at the given wavelength of the laser beam. We can use the table values for the refractive index at various wavelengths, or we can use the Cauchy formula to approximate the refractive index of water with wavelength:
n(water) = A + B/λ^2 + C/λ^4
Where A, B, and C are constants and λ is the wavelength. For visible light, we can use an approximate wavelength of 550 nm.
Assuming A = 1.33, B = 0.0013, and C = 0, we can calculate:
n(water) = 1.33 + 0.0013/(550^2) ≈ 1.333
Now, we can estimate the refractive index of the glucose solution by assuming it is proportional to the concentration. Since the given solution has a 25% glucose concentration, we can estimate the refractive index using the formula:
n2 ≈ n(water) + (concentration/100) * k
Where n(water) is the refractive index of water, concentration is the percentage of glucose, and k is a proportionality constant. Assuming k ≈ 0.001, we can calculate:
n2 ≈ 1.333 + (25/100) * 0.001 ≈ 1.336
Now that we have the refractive indices of both air (n1 = 1) and the glucose solution (n2 ≈ 1.336), we can apply Snell's law to find the angle of refraction (θ2):
n1*sin(θ1) = n2*sin(θ2)
1*sin(37°) = 1.336*sin(θ2)
sin(θ2) = (1*sin(37°)) / 1.336
θ2 ≈ arcsin((1*sin(37°)) / 1.336)
Using a calculator, we find θ2 ≈ 26.98°.
Therefore, the light will travel in the glucose solution at an angle of approximately 26.98° with respect to the normal.