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. `1 : mu`
B. `1 : mu^(2)`
C. `mu : 1`
D. `1 : 1`

Answer

To solve the problem of finding 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 vacuum (or air) to the speed of light in the medium. For air, the refractive index is approximately 1, and for glass, it is given as μ.

### Step 2: Relate the wavelengths in different media
When light travels from one medium to another, its speed and wavelength change, but its frequency remains constant. The relationship between the speed of light (c), wavelength (λ), and frequency (f) is given by:
\[ c = f \cdot \lambda \]

### Step 3: Write the equations for both media
Let:
- \( \lambda_1 \) = wavelength of light in air
- \( \lambda_2 \) = wavelength of light in glass

In air:
\[ c = f \cdot \lambda_1 \]

In glass:
\[ \frac{c}{\mu} = f \cdot \lambda_2 \]

### Step 4: Express the speed of light in terms of wavelength
From the first equation, we can express frequency as:
\[ f = \frac{c}{\lambda_1} \]

Substituting this into the second equation gives:
\[ \frac{c}{\mu} = \frac{c}{\lambda_1} \cdot \lambda_2 \]

### Step 5: Simplify the equation
Cancelling \( c \) from both sides (assuming \( c \neq 0 \)):
\[ \frac{1}{\mu} = \frac{\lambda_2}{\lambda_1} \]

### Step 6: Rearranging to find the ratio of wavelengths
Rearranging the equation gives:
\[ \frac{\lambda_1}{\lambda_2} = \mu \]

### Conclusion
The ratio of the wavelength of incident light in air to the wavelength of refracted light in glass is:
\[ \frac{\lambda_1}{\lambda_2} = \mu \]

### Final Answer
Thus, the ratio of the wavelength of incident and refracted waves is \( \mu : 1 \).

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