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A glass plate P (refractive index np = 1.54) is coated with a dielectric material C with the refractive index nc = 1.6. In order to have enhanced reflection from this coated glass for near normal incident light of wavelength λ, the thickness of the coating material. C must be
- a)even multiples of λ/2nc
- b)even multiples of λ/4nc
- c)odd multiples of λ/4nc
- d)integral multiples of λ/4nc
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A glass plate P (refractive index np = 1.54) is coated with a dielectr...
For enhanced reflection path difference in the coating must be an integral multiple of λ.
θ = 0 For near normal incident ray
θ = 0 For near normal incident ray
Most Upvoted Answer
A glass plate P (refractive index np = 1.54) is coated with a dielectr...
To enhance reflection from the coated glass plate, we need to create a high contrast in refractive indices between the glass plate and the dielectric coating. This can be achieved by adjusting the thickness of the coating.
The condition for enhanced reflection from a thin film is when the phase change upon reflection from the top surface and the bottom surface of the film is equal. This condition is known as the constructive interference condition.
For near normal incident light, the condition for constructive interference can be expressed as:
2ncd = mλ
Where:
nc is the refractive index of the coating (1.6)
d is the thickness of the coating
m is an integer representing the order of the interference (m = 0, 1, 2, ...)
Since we want enhanced reflection, we want to maximize the intensity of the reflected light. For this, we generally want the first-order (m = 1) interference condition, as it gives the strongest reflection.
Therefore, for the first-order constructive interference condition:
2ncd = λ
Rearranging the equation:
d = λ / (2nc)
Substituting the values:
λ = wavelength of the incident light
Let's assume the incident light has a wavelength of 600 nm (0.6 μm).
d = 0.6 μm / (2 * 1.6)
d ≈ 0.1875 μm
Therefore, to have enhanced reflection for near-normal incident light with a wavelength of 600 nm, the thickness of the dielectric coating should be approximately 0.1875 μm.
The condition for enhanced reflection from a thin film is when the phase change upon reflection from the top surface and the bottom surface of the film is equal. This condition is known as the constructive interference condition.
For near normal incident light, the condition for constructive interference can be expressed as:
2ncd = mλ
Where:
nc is the refractive index of the coating (1.6)
d is the thickness of the coating
m is an integer representing the order of the interference (m = 0, 1, 2, ...)
Since we want enhanced reflection, we want to maximize the intensity of the reflected light. For this, we generally want the first-order (m = 1) interference condition, as it gives the strongest reflection.
Therefore, for the first-order constructive interference condition:
2ncd = λ
Rearranging the equation:
d = λ / (2nc)
Substituting the values:
λ = wavelength of the incident light
Let's assume the incident light has a wavelength of 600 nm (0.6 μm).
d = 0.6 μm / (2 * 1.6)
d ≈ 0.1875 μm
Therefore, to have enhanced reflection for near-normal incident light with a wavelength of 600 nm, the thickness of the dielectric coating should be approximately 0.1875 μm.
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A glass plate P (refractive index np = 1.54) is coated with a dielectric material C with the refractive index nc = 1.6. In order to have enhanced reflection from this coated glass for near normal incident light of wavelength λ, the thickness of the coating material. C must bea)even multiples ofλ/2ncb)even multiples ofλ/4ncc)odd multiples ofλ/4ncd)integral multiples ofλ/4ncCorrect answer is option 'B'. Can you explain this answer?
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