Monochromatic light is refracted from air into the glass of refractive index `mu` . The ratio of the wavelength of incident and refracted waves is

Options

A. (a)`1:mu`
B. (b)`1:mu^(2)`
C. (c)`mu:1`
D. (d)`1:1`

Answer

To find the ratio of the wavelength of incident and refracted waves when monochromatic light is refracted from air into glass, we can follow these steps:

### Step 1: Understand the refractive index
The refractive index (μ) of a medium is defined as the ratio of the speed of light in a vacuum (or air, approximately) to the speed of light in that medium. For air, we can consider the refractive index to be approximately 1. Therefore, we have:
- Refractive index of air (n_air) = 1
- Refractive index of glass (n_glass) = μ

### Step 2: Apply Snell's Law
According to Snell's Law, the relationship between the angles of incidence (i) and refraction (r) is given by:
\[ \frac{\sin(i)}{\sin(r)} = \frac{n_{\text{air}}}{n_{\text{glass}}} = \frac{1}{\mu} \]

### Step 3: Relate speed and wavelength
The speed of light in a medium is related to its wavelength. The speed of light in air (v1) and in glass (v2) can be expressed in terms of their respective wavelengths (λ1 for air and λ2 for glass):
\[ v_1 = f \cdot \lambda_1 \]
\[ v_2 = f \cdot \lambda_2 \]
where f is the frequency of the light, which remains constant when light travels from one medium to another.

### Step 4: Use the relationship between speed and refractive index
The refractive index can also be expressed in terms of the speeds of light in the two media:
\[ \mu = \frac{v_1}{v_2} \]
Substituting the expressions for v1 and v2, we get:
\[ \mu = \frac{f \cdot \lambda_1}{f \cdot \lambda_2} = \frac{\lambda_1}{\lambda_2} \]

### Step 5: Find the ratio of wavelengths
From the above equation, we can rearrange it to find the ratio of the wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \mu \]
This means that the ratio of the wavelength of the incident light (in air) to the wavelength of the refracted light (in glass) is equal to the refractive index of the glass.

### Final Result
Thus, the ratio of the wavelength of the incident wave to the refracted wave is:
\[ \frac{\lambda_1}{\lambda_2} = \mu : 1 \]

### Conclusion
The correct answer is that the ratio of the wavelength of incident and refracted waves is μ : 1.

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