The critical angle for refraction from medium -1 to air is `theta_(1)` and that from medium _2 to air is `theta_(2)` . If medium _2 is denser than medium _1. The critical angle for refraction from medium _2 to medium _ 1 is

Options

A. `sin^(-1)((sintheta_(2))/(sintheta_(1)))`
B. `sin^(-1)((sintheta_(1))/(sintheta_(2)))`
C. `sin^(-1)(sintheta_(2))`
D. `sin^(-1)(sintheta_(1))`

Explanation

To find the critical angle for refraction from medium 2 to medium 1, we will use Snell's law and the definitions of critical angles. Here is a step-by-step solution:

### Step 1: Understanding Critical Angle
The critical angle is defined as the angle of incidence above which total internal reflection occurs. For light traveling from a denser medium to a less dense medium (like air), it can be calculated using Snell's law.

### Step 2: Applying Snell's Law for Medium 1 to Air
For medium 1 (with refractive index μ1) to air (with refractive index μ_air = 1), the critical angle θ1 can be expressed as:
\[ \mu_1 \sin(\theta_1) = \mu_{\text{air}} \sin(90^\circ) \]
Since \(\sin(90^\circ) = 1\), we have:
\[ \mu_1 \sin(\theta_1) = 1 \]
Thus,
\[ \sin(\theta_1) = \frac{1}{\mu_1} \]

### Step 3: Applying Snell's Law for Medium 2 to Air
For medium 2 (with refractive index μ2) to air, the critical angle θ2 can be expressed similarly:
\[ \mu_2 \sin(\theta_2) = \mu_{\text{air}} \sin(90^\circ) \]
So,
\[ \mu_2 \sin(\theta_2) = 1 \]
This gives us:
\[ \sin(\theta_2) = \frac{1}{\mu_2} \]

### Step 4: Finding the Critical Angle from Medium 2 to Medium 1
Now, we want to find the critical angle (let's call it α) for the transition from medium 2 to medium 1. According to Snell's law:
\[ \mu_2 \sin(\alpha) = \mu_1 \sin(90^\circ) \]
This simplifies to:
\[ \mu_2 \sin(\alpha) = \mu_1 \]
Thus,
\[ \sin(\alpha) = \frac{\mu_1}{\mu_2} \]

### Step 5: Substituting Values of μ1 and μ2
From our earlier steps, we have:
\[ \mu_1 = \frac{1}{\sin(\theta_1)} \]
and
\[ \mu_2 = \frac{1}{\sin(\theta_2)} \]
Substituting these into the equation for sin(α):
\[ \sin(\alpha) = \frac{\frac{1}{\sin(\theta_1)}}{\frac{1}{\sin(\theta_2)}} = \frac{\sin(\theta_2)}{\sin(\theta_1)} \]

### Step 6: Finding α
To find α, we take the inverse sine:
\[ \alpha = \sin^{-1}\left(\frac{\sin(\theta_2)}{\sin(\theta_1)}\right) \]

### Conclusion
Thus, the critical angle for refraction from medium 2 to medium 1 is given by:
\[ \alpha = \sin^{-1}\left(\frac{\sin(\theta_2)}{\sin(\theta_1)}\right) \]