What is the equation for calculating the amount of light that gets lost via reflection for a spectacle lens?

Study for the NBEO Physiological Optics Test. Prepare with flashcards and multiple-choice questions, each question has hints and explanations. Ace your exam!

Multiple Choice

What is the equation for calculating the amount of light that gets lost via reflection for a spectacle lens?

Explanation:
Light that hits a boundary between two materials can be partly reflected, and the amount reflected depends on the refractive indices on either side. At normal incidence, the fraction of light reflected, R, is determined by the Fresnel relation R = ((n2 − n1)/(n2 + n1))^2. This result comes from applying the boundary conditions for the electric field at the interface and looking at the reflected amplitude relative to the incident one; when you square the amplitude ratio to get intensity, you obtain this expression for reflectance. The sign in the numerator doesn’t matter after squaring, so using (n1 − n2)/(n1 + n2) gives the same R. In practice, with air (n1 ≈ 1.0) and common lens material (n2 ≈ 1.5), you’d get about ((1.5 − 1.0)/(1.5 + 1.0))^2 ≈ (0.5/2.5)^2 ≈ 0.04, or 4% of the light reflected per surface. This is why anti-reflective coatings are beneficial for spectacles.

Light that hits a boundary between two materials can be partly reflected, and the amount reflected depends on the refractive indices on either side. At normal incidence, the fraction of light reflected, R, is determined by the Fresnel relation R = ((n2 − n1)/(n2 + n1))^2. This result comes from applying the boundary conditions for the electric field at the interface and looking at the reflected amplitude relative to the incident one; when you square the amplitude ratio to get intensity, you obtain this expression for reflectance. The sign in the numerator doesn’t matter after squaring, so using (n1 − n2)/(n1 + n2) gives the same R. In practice, with air (n1 ≈ 1.0) and common lens material (n2 ≈ 1.5), you’d get about ((1.5 − 1.0)/(1.5 + 1.0))^2 ≈ (0.5/2.5)^2 ≈ 0.04, or 4% of the light reflected per surface. This is why anti-reflective coatings are beneficial for spectacles.

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