White light is sent through an interface of a 100% (w/v) glycerol solution (n1 = 1.474) and a 20% (w/v) sucrose solution (n2=1.364) At an angle of: A) Theta=33 degree, determine the angle of Theta2 in degrees () B) Theta 1 =0degree, determine the angle or Theta2 in degrees () A) Theta2= Number degree B) Theta2= Number degree
The Correct Answer and Explanation is:
Let’s analyze the problem step-by-step:
Given data:
- Medium 1 (glycerol solution): refractive index $n_1 = 1.474$
- Medium 2 (sucrose solution): refractive index $n_2 = 1.364$
- Angle of incidence $\theta_1 = 33^\circ$ for part A
- Angle of incidence $\theta_1 = 0^\circ$ for part B
Task:
- Find the angle of refraction $\theta_2$ in degrees for both cases.
Key principle:
Snell’s Law governs refraction at an interface between two media with different refractive indices:
$$
n_1 \sin \theta_1 = n_2 \sin \theta_2
$$
Part A: $\theta_1 = 33^\circ$
$$
n_1 \sin \theta_1 = n_2 \sin \theta_2
$$
$$
1.474 \times \sin(33^\circ) = 1.364 \times \sin \theta_2
$$
Calculate $\sin(33^\circ)$:
$$
\sin(33^\circ) \approx 0.5446
$$
Substitute:
$$
1.474 \times 0.5446 = 1.364 \times \sin \theta_2
$$
$$
0.8027 = 1.364 \times \sin \theta_2
$$
Solve for $\sin \theta_2$:
$$
\sin \theta_2 = \frac{0.8027}{1.364} \approx 0.5885
$$
Find $\theta_2$:
$$
\theta_2 = \arcsin(0.5885) \approx 36.0^\circ
$$
Part B: $\theta_1 = 0^\circ$
At normal incidence, light hits the interface head-on.
Apply Snell’s Law:
$$
n_1 \sin 0^\circ = n_2 \sin \theta_2
$$
Since $\sin 0^\circ = 0$:
$$
0 = n_2 \sin \theta_2 \implies \sin \theta_2 = 0 \implies \theta_2 = 0^\circ
$$
Final answers:
- A) $\theta_2 = 36.0^\circ$
- B) $\theta_2 = 0^\circ$
Explanation (300+ words):
When light passes from one medium to another with a different refractive index, its speed changes, which causes the light to bend or refract. The relationship between the angle of incidence $\theta_1$ (the angle the incoming light makes with the normal) and the angle of refraction $\theta_2$ (the angle the refracted light makes with the normal) is described by Snell’s Law:
$$
n_1 \sin \theta_1 = n_2 \sin \theta_2
$$
where $n_1$ and $n_2$ are the refractive indices of the first and second media, respectively.
In this problem, white light travels from a 100% glycerol solution (higher refractive index $n_1 = 1.474$) into a 20% sucrose solution (lower refractive index $n_2 = 1.364$). Since the refractive index of glycerol is greater than that of sucrose, light will bend away from the normal upon entering the sucrose solution.
For part A, the incident angle is $33^\circ$. Applying Snell’s Law, the sine of the refracted angle is calculated to be approximately 0.5885, which corresponds to an angle of about $36.0^\circ$. This means the light bends slightly away from the normal as it passes into the sucrose solution, consistent with the lower refractive index.
For part B, the light hits the interface at normal incidence ($0^\circ$). At normal incidence, there is no bending because the light travels straight through the interface. Snell’s Law simplifies to zero on both sides, and the refracted angle remains $0^\circ$.
This example demonstrates the fundamental optical principle that when light moves from a denser medium to a less dense medium (higher to lower refractive index), the refracted ray bends away from the normal, and the refraction angle is larger than the incidence angle. Conversely, if the incidence is zero (straight on), no bending occurs