Solving Thin Film Problem: Wavelength in Visible Spectrum

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To determine the wavelength most strongly transmitted through a magnesium fluoride film on a camera lens, the correct approach involves using the formula for destructive interference, 2t = m(λ/n), where t is the film thickness, m is an integer, and λ is the wavelength in the medium. The index of refraction of the film (n = 1.45) is crucial in calculating the effective wavelength. The correct wavelength in the visible spectrum that results in maximum transmission is found to be around 640 nm, which corresponds to red light. This wavelength is selected because it minimizes reflection at the first surface of the film. Understanding the role of the film's refractive index is essential for solving similar problems effectively.
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A camera lens (n = 1.80) is coated with a thin film of magnesium fluoride (n = 1.45) of thickness 80.0 nm. What wavelength in the visible spectrum is most strongly transmitted through the film?

I have tried to solve this problem for quite a while now, and i hate to admit that there might be a small thing that i am missing. i have reached the following answers : 116 nm, 160 nm and 232 nm and all of them are wrong. I am using this formula: 2t= m*(lambda). I only have 2 tries left so please help me out. Also include a reason why you reached that answer. I have a final comming up and i would love to learn the propper way to do this. Thank you in advance.
 
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You want destructive interference for reflection at the first surface.
I think this means lambda in the glass=t.
 
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Thanks for your attempt, but I already figured out the correct way of doing it. The index of refraction of the MgF2 is the key to the answer.
 
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