Find the critical angle for liquid-air interface if the refractive index of the liquid is 2.30°45°60°25°

To find the critical angle for the liquid-air interface, we can use Snell's Law, which states:

$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$

Where:

• $n_1$ is the refractive index of the liquid,
• $n_2$ is the refractive index of air (approximately 1.0),
• $\theta_1$ is the angle of incidence (the critical angle in this case),
• $\theta_2$ is the angle of refraction (which will be $90^\circ$ at the critical angle).

Step 1: Break Down the Problem

1. Identify the values:
   • $n_1 = 2.30$
   • $n_2 = 1.0$
2. Identify the conditions for the critical angle.

Step 2: Relevant Concepts

Using Snell's Law for the critical angle: $n_1 \sin(\theta_c) = n_2 \sin(90^\circ)$ Here, since $\sin(90^\circ) = 1$, it simplifies to: $n_1 \sin(\theta_c) = n_2$

Step 3: Analysis and Detail

Rearranging the equation gives us: $\sin(\theta_c) = \frac{n_2}{n_1} = \frac{1.0}{2.30}$ Calculating the value: $\sin(\theta_c) = \frac{1}{2.30} \approx 0.4348$

To find the critical angle $\theta_c$: $\theta_c = \arcsin(0.4348)$

Using a calculator, $\theta_c \approx 25.8^\circ$

Step 4: Verify and Summarize

We've calculated: $\sin(\theta_c) \approx 0.4348 \Rightarrow \theta_c \approx 25.8^\circ$

Final Answer

The critical angle for the liquid-air interface is approximately $25.8^\circ$.