Wave length of light in denser mediumis `4000Å`, it is grazing into a rarer medium. If critical angle for the pairof mediam is `sin^(-1)(2/3)` then the wave length of light in rarer medium is

Options

A. `4000Å`
B. `2666Å`
C. `8000Å`
D. `6000Å`

Answer

To find the wavelength of light in the rarer medium, we can follow these steps:

### Step 1: Understand the relationship between critical angle and refractive index
The critical angle (ic) is related to the refractive indices of the two media by the formula:
\[ \sin(i_c) = \frac{1}{\mu} \]
where \(\mu\) is the refractive index of the denser medium with respect to the rarer medium.

### Step 2: Calculate the refractive index
Given that the critical angle \(i_c = \sin^{-1}\left(\frac{2}{3}\right)\), we can find the refractive index:
\[ \sin(i_c) = \frac{2}{3} \implies \frac{1}{\mu} = \frac{2}{3} \implies \mu = \frac{3}{2} \]

### Step 3: Use the relationship between wavelengths and refractive index
The relationship between the wavelengths in the two media is given by:
\[ \mu = \frac{\lambda_2}{\lambda_1} \]
where \(\lambda_1\) is the wavelength in the denser medium and \(\lambda_2\) is the wavelength in the rarer medium.

### Step 4: Substitute the known values
We know that \(\lambda_1 = 4000 \, \text{Å}\) (in the denser medium) and \(\mu = \frac{3}{2}\):
\[ \frac{3}{2} = \frac{\lambda_2}{4000} \]

### Step 5: Solve for \(\lambda_2\)
Now, we can rearrange the equation to find \(\lambda_2\):
\[ \lambda_2 = \frac{3}{2} \times 4000 \]
\[ \lambda_2 = 6000 \, \text{Å} \]

### Final Answer
The wavelength of light in the rarer medium is \(6000 \, \text{Å}\).

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