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When white light is incident normally an...

When white light is incident normally an oil film of thickness `10^-4 cm` and refractive index. 1.4 then the wavelength which will not be seen in the reflected system of light is

A

7000A

B

5600A

C

4000A

D

All of these

Text Solution

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The correct Answer is:
To solve the problem of finding the wavelength of light that will not be seen in the reflected system due to destructive interference in an oil film, we can follow these steps: ### Step-by-Step Solution: 1. **Identify Given Values**: - Thickness of the oil film, \( T = 10^{-4} \, \text{cm} = 10^{-6} \, \text{m} \) - Refractive index of oil, \( \mu = 1.4 \) 2. **Determine the Phase Change**: - When light reflects off a denser medium (from air to oil), it undergoes a phase change of \( \pi \) (or half a wavelength). - The light reflecting off the top surface (air to oil) does not undergo a phase change. 3. **Calculate the Path Difference**: - The path difference for light reflecting from the oil film is given by: \[ \text{Path Difference} = 2 \mu T \] - Substituting the values: \[ \text{Path Difference} = 2 \times 1.4 \times 10^{-6} \, \text{m} = 2.8 \times 10^{-6} \, \text{m} \] 4. **Condition for Destructive Interference**: - For destructive interference, the condition is: \[ 2 \mu T = (n + \frac{1}{2}) \lambda \] - Rearranging gives: \[ \lambda = \frac{2 \mu T}{n + \frac{1}{2}} \] 5. **Calculate Wavelengths for Different Values of \( n \)**: - For \( n = 0 \): \[ \lambda_0 = \frac{2.8 \times 10^{-6}}{0 + \frac{1}{2}} = 5.6 \times 10^{-6} \, \text{m} = 5600 \, \text{Å} \] - For \( n = 1 \): \[ \lambda_1 = \frac{2.8 \times 10^{-6}}{1 + \frac{1}{2}} = \frac{2.8 \times 10^{-6}}{1.5} = 1.8667 \times 10^{-6} \, \text{m} = 18667 \, \text{Å} \] - For \( n = 2 \): \[ \lambda_2 = \frac{2.8 \times 10^{-6}}{2 + \frac{1}{2}} = \frac{2.8 \times 10^{-6}}{2.5} = 1.12 \times 10^{-6} \, \text{m} = 11200 \, \text{Å} \] - For \( n = 3 \): \[ \lambda_3 = \frac{2.8 \times 10^{-6}}{3 + \frac{1}{2}} = \frac{2.8 \times 10^{-6}}{3.5} = 0.8 \times 10^{-6} \, \text{m} = 8000 \, \text{Å} \] - For \( n = 4 \): \[ \lambda_4 = \frac{2.8 \times 10^{-6}}{4 + \frac{1}{2}} = \frac{2.8 \times 10^{-6}}{4.5} = 0.6222 \times 10^{-6} \, \text{m} = 6222 \, \text{Å} \] 6. **Conclusion**: - The wavelengths that will not be seen in the reflected light due to destructive interference are \( 5600 \, \text{Å} \), \( 8000 \, \text{Å} \), and \( 6222 \, \text{Å} \). The correct answer is \( 5600 \, \text{Å} \).
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