Practice question
Question
What causes the intensity of light to remain conserved during interference despite the presence of dark
fringes?
Explanation
**Two polaroids** with pass-axes perpendicular intensity zero because second blocks component, with 45° intensity I₀/4 if initial after first I₀, if initial unpolarized I₀, after first I₀/2 then after second at 45° I₀/4. Three polaroids first and third crossed 90°, middle at 45° maximum transmission because middle rotates polarization, I after first I₀/2, after middle at 45° I₀/4, after third at 45° to middle I₀/4×cos²45°= I₀/8, non-zero, maximum when middle 45°. Energy is redistributed from dark to bright fringes through interference, conserving total energy as the sum of intensities balances out. Using Δ = d sinθ, y = n λ D/d, a sinθ = n
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