White light consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen , When white light is passed through glass of refractive index 1.55 ?

Answer

To find the wavelength range of white light when it is passed through a glass medium with a refractive index of 1.55, we can follow these steps:

### Step 1: Understand the relationship between wavelength in vacuum and in a medium
The wavelength of light changes when it passes from one medium to another. The relationship is given by the formula:
\[ \lambda' = \frac{\lambda}{\mu} \]
where:
- \(\lambda'\) is the wavelength in the medium,
- \(\lambda\) is the wavelength in vacuum (or air),
- \(\mu\) is the refractive index of the medium.

### Step 2: Identify the range of wavelengths in white light
White light consists of wavelengths ranging from 400 nm to 700 nm. Therefore:
- Minimum wavelength (\(\lambda_{min}\)) = 400 nm
- Maximum wavelength (\(\lambda_{max}\)) = 700 nm

### Step 3: Calculate the new wavelength for the minimum wavelength (400 nm)
Using the formula from Step 1:
\[ \lambda'_{min} = \frac{400 \text{ nm}}{1.55} \]
Calculating this gives:
\[ \lambda'_{min} = \frac{400}{1.55} \approx 258.06 \text{ nm} \]

### Step 4: Calculate the new wavelength for the maximum wavelength (700 nm)
Using the same formula:
\[ \lambda'_{max} = \frac{700 \text{ nm}}{1.55} \]
Calculating this gives:
\[ \lambda'_{max} = \frac{700}{1.55} \approx 451.61 \text{ nm} \]

### Step 5: State the final wavelength range
Thus, when white light is passed through glass with a refractive index of 1.55, the wavelength range seen will be:
\[ \text{Wavelength range} = 258.06 \text{ nm} \text{ to } 451.61 \text{ nm} \]

### Summary
The final wavelength range of white light when passed through glass of refractive index 1.55 is approximately 258.06 nm to 451.61 nm.

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