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A camera lens in air (index 1.00) is coated with a thin film with a thickness L that maximizes the transmitted light (and minimizes the reflected light) for a particular wavelength. If the thin film has an index of 1.25 and the lens material an index of 1.50, what minimum thickness L is required? (Let λn be the wavelength in the thin film.)

Respuesta :

Answer:

t = 100 nm

Explanation:

given,

Refractive index of coating is n ' = 1.25

refractive index of glass is n = 1.5

assuming wavelength of the film = 500 nm

destructive interference is

[tex]m +\dfrac{1}{2}\times \lambda = 2 n't [/tex]

[tex]t = \dfrac{m +\dfrac{1}{2}\times \lambda }{2 n'}[/tex]

for minimum thickness m = 0

[tex]t = \dfrac{0+\dfrac{1}{2}\times \lambda}{2 n'}[/tex]

[tex]t = \dfrac{500}{4 n'}[/tex]

[tex]t = \dfrac{500}{4\times 1.25}[/tex]

      t = 100 nm

hence, the minimum thickness required is t = 100 nm

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