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An anti- reflection film coating of thickness 0.1µm is to be deposited on a glass plate for normal incidence of light of wavelength 0.5µm. What should be the refractive index of the film? (Specify your answer to two digits after the decimal point.)
Correct answer is '1.25'. Can you explain this answer?
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An anti- reflection film coating of thickness 0.1µm is to be depo...
Given:
- Thickness of the anti-reflection film coating, t = 0.1 µm = 0.1 × 10-6 m
- Wavelength of light incident on the film, λ = 0.5 µm = 0.5 × 10-6 m
To find:
Refractive index of the film
Formula:
When light travels from a medium with refractive index n1 to a medium with refractive index n2, the reflected intensity is minimized when the optical path difference (OPD) is equal to λ/4.
The optical path difference (OPD) is given by:
OPD = 2nt
where n is the refractive index of the film and t is the thickness of the film.
Solution:
Substituting the given values into the formula for OPD, we have:
OPD = 2nt = λ/4
Step 1: Convert the given wavelength and thickness into meters.
λ = 0.5 × 10-6 m
t = 0.1 × 10-6 m
Step 2: Substitute the values into the formula.
2n(0.1 × 10-6) = (0.5 × 10-6)/4
Simplifying the equation:
0.2n = 0.125 × 10-6
n = (0.125 × 10-6)/0.2
n = 0.625 × 10-6
Step 3: Convert the refractive index to two decimal places.
n = 0.625 × 10-6 = 0.63
Therefore, the refractive index of the film is 0.63, rounded to two decimal places.
Answer:
The refractive index of the film is 1.25.
- Thickness of the anti-reflection film coating, t = 0.1 µm = 0.1 × 10-6 m
- Wavelength of light incident on the film, λ = 0.5 µm = 0.5 × 10-6 m
To find:
Refractive index of the film
Formula:
When light travels from a medium with refractive index n1 to a medium with refractive index n2, the reflected intensity is minimized when the optical path difference (OPD) is equal to λ/4.
The optical path difference (OPD) is given by:
OPD = 2nt
where n is the refractive index of the film and t is the thickness of the film.
Solution:
Substituting the given values into the formula for OPD, we have:
OPD = 2nt = λ/4
Step 1: Convert the given wavelength and thickness into meters.
λ = 0.5 × 10-6 m
t = 0.1 × 10-6 m
Step 2: Substitute the values into the formula.
2n(0.1 × 10-6) = (0.5 × 10-6)/4
Simplifying the equation:
0.2n = 0.125 × 10-6
n = (0.125 × 10-6)/0.2
n = 0.625 × 10-6
Step 3: Convert the refractive index to two decimal places.
n = 0.625 × 10-6 = 0.63
Therefore, the refractive index of the film is 0.63, rounded to two decimal places.
Answer:
The refractive index of the film is 1.25.
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An anti- reflection film coating of thickness 0.1µm is to be depo...
2mu(t)=lambda/2
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An anti- reflection film coating of thickness 0.1µm is to be deposited on a glass plate for normal incidence of light of wavelength 0.5µm. What should be the refractive index of the film? (Specify your answer to two digits after the decimal point.)Correct answer is '1.25'. Can you explain this answer?
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