When a light ray is refracted from one medium into another, the wavelength changes from `4500overset(o)(A)` to `3600overset(o)(A)`. The critial angle for a ray from second medium to first medium is

Options

A. `sin^(-1)(2/13)`
B. `cos^(-1)(2/3)`
C. `tan^(-1)(3/2)`
D. `tan^(-1)(2/sqrt(5))`

Explanation

To solve the problem, we need to find the critical angle when a light ray is refracted from the second medium back into the first medium. Here are the steps to derive the solution:

### Step 1: Understand the relationship between wavelength and refractive index
When light travels from one medium to another, its wavelength changes according to the refractive indices of the two media. The relationship can be expressed as:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\mu_2}{\mu_1} \]
where \(\lambda_1\) is the wavelength in the first medium, \(\lambda_2\) is the wavelength in the second medium, \(\mu_1\) is the refractive index of the first medium, and \(\mu_2\) is the refractive index of the second medium.

### Step 2: Substitute the given values
We are given:
- \(\lambda_1 = 4500 \, \text{Å}\)
- \(\lambda_2 = 3600 \, \text{Å}\)

Using the relationship from Step 1, we can find the ratio of the refractive indices:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\mu_2}{\mu_1} \]
Substituting the values:
\[ \frac{4500}{3600} = \frac{\mu_2}{\mu_1} \]

### Step 3: Simplify the ratio
Calculating the ratio:
\[ \frac{4500}{3600} = \frac{45}{36} = \frac{5}{4} \]
Thus, we have:
\[ \frac{\mu_2}{\mu_1} = \frac{5}{4} \]
Taking the reciprocal gives us:
\[ \frac{\mu_1}{\mu_2} = \frac{4}{5} \]

### Step 4: Use the critical angle formula
The critical angle \(\theta_c\) can be calculated using Snell's law:
\[ \sin \theta_c = \frac{\mu_1}{\mu_2} \]
Substituting the value we found:
\[ \sin \theta_c = \frac{4}{5} \]

### Step 5: Calculate the critical angle
To find \(\theta_c\), we take the inverse sine:
\[ \theta_c = \sin^{-1}\left(\frac{4}{5}\right) \]

### Final Answer
Thus, the critical angle for the ray coming from the second medium to the first medium is:
\[ \theta_c = \sin^{-1}\left(\frac{4}{5}\right) \]