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A light of wavelength 5890Å falls normal...

A light of wavelength `5890Å` falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is

A

`2.945xx10^(-7)m`

B

`3.945xx10^(-7)m`

C

`4.95xx10^(-7)m`

D

`1.945xx10^(-7)m`

Text Solution

AI Generated Solution

The correct Answer is:
To find the minimum thickness of a thin air film such that it appears dark in reflected light when illuminated by light of wavelength \(5890 \, \text{Å}\), we can follow these steps: ### Step 1: Understand the Condition for Dark Appearance For a thin film to appear dark in reflected light, we need to consider the interference of light waves. The condition for destructive interference (dark appearance) in a thin film is given by the formula: \[ 2 \mu T = (n + \frac{1}{2}) \lambda \] where: - \( \mu \) is the refractive index of the film, - \( T \) is the thickness of the film, - \( n \) is an integer (0, 1, 2, ...), - \( \lambda \) is the wavelength of light. ### Step 2: Identify Parameters In this case: - The wavelength of light \( \lambda = 5890 \, \text{Å} = 5890 \times 10^{-10} \, \text{m} \). - The refractive index of air \( \mu \approx 1 \). - For minimum thickness, we take \( n = 0 \). ### Step 3: Substitute Values into the Formula Using the condition for destructive interference and substituting \( n = 0 \): \[ 2 \mu T = (0 + \frac{1}{2}) \lambda \] This simplifies to: \[ 2 \mu T = \frac{1}{2} \lambda \] ### Step 4: Solve for Thickness \( T \) Now, substituting \( \mu = 1 \) and rearranging for \( T \): \[ T = \frac{\lambda}{4 \mu} \] Substituting the values: \[ T = \frac{5890 \times 10^{-10}}{4 \times 1} = \frac{5890 \times 10^{-10}}{4} \] Calculating this gives: \[ T = \frac{5890 \times 10^{-10}}{4} = 1472.5 \times 10^{-10} \, \text{m} = 1.4725 \times 10^{-7} \, \text{m} \] ### Step 5: Convert to Appropriate Units To express the thickness in meters: \[ T = 1.4725 \times 10^{-7} \, \text{m} = 147.25 \, \text{nm} \] ### Final Answer Thus, the minimum thickness of the film such that it appears dark in reflected light is: \[ \boxed{1.4725 \times 10^{-7} \, \text{m}} \]
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