A thin film of oil (n = 1.5) is on water (n = 1.33). What is the condition for destructive interference for light of wavelength 600 nm in air? (2022)
Practice Questions
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Q1
A thin film of oil (n = 1.5) is on water (n = 1.33). What is the condition for destructive interference for light of wavelength 600 nm in air? (2022)
2t = (m + 1/2)λ
2t = mλ
2t = (m + 1)λ
2t = (m - 1/2)λ
For destructive interference in a thin film with a higher refractive index below, the condition is 2t = (m + 1/2)λ/n, where n is the refractive index of the film.
Questions & Step-by-step Solutions
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Q: A thin film of oil (n = 1.5) is on water (n = 1.33). What is the condition for destructive interference for light of wavelength 600 nm in air? (2022)
Solution: For destructive interference in a thin film with a higher refractive index below, the condition is 2t = (m + 1/2)λ/n, where n is the refractive index of the film.
Steps: 6
Step 1: Identify the refractive indices of the materials involved. The oil has a refractive index (n) of 1.5 and the water has a refractive index of 1.33.
Step 2: Understand that for destructive interference in a thin film, the light reflecting off the top surface and the light reflecting off the bottom surface must be out of phase.
Step 3: Recognize that when light reflects off a medium with a higher refractive index (oil to air), it undergoes a phase change of 180 degrees (or half a wavelength).
Step 4: Use the formula for destructive interference in a thin film: 2t = (m + 1/2)λ/n, where 't' is the thickness of the film, 'm' is an integer (0, 1, 2,...), 'λ' is the wavelength of light in air, and 'n' is the refractive index of the film (oil in this case).
Step 5: Substitute the values into the formula. Here, λ = 600 nm and n = 1.5.
Step 6: Rearrange the formula to find the condition for destructive interference: 2t = (m + 1/2)(600 nm) / 1.5.
