logistic_guy Senior Member Joined Apr 17, 2024 Messages 2,213 Jun 29, 2025 #1 At what angle will 440\displaystyle 440440-nm\displaystyle \text{nm}nm light produce a third-order maximum when falling on a grating whose slits are 1.35×10−3 cm\displaystyle 1.35 \times 10^{-3} \ \text{cm}1.35×10−3 cm apart?
At what angle will 440\displaystyle 440440-nm\displaystyle \text{nm}nm light produce a third-order maximum when falling on a grating whose slits are 1.35×10−3 cm\displaystyle 1.35 \times 10^{-3} \ \text{cm}1.35×10−3 cm apart?
K khansaheb Senior Member Joined Apr 6, 2023 Messages 1,227 Jun 30, 2025 #2 logistic_guy said: At what angle will 440\displaystyle 440440-nm\displaystyle \text{nm}nm light produce a third-order maximum when falling on a grating whose slits are 1.35×10−3 cm\displaystyle 1.35 \times 10^{-3} \ \text{cm}1.35×10−3 cm apart? Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem
logistic_guy said: At what angle will 440\displaystyle 440440-nm\displaystyle \text{nm}nm light produce a third-order maximum when falling on a grating whose slits are 1.35×10−3 cm\displaystyle 1.35 \times 10^{-3} \ \text{cm}1.35×10−3 cm apart? Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem
logistic_guy Senior Member Joined Apr 17, 2024 Messages 2,213 Jul 8, 2025 #3 We use the diffraction maxima formula. dsinθ=mλ\displaystyle d\sin \theta = m\lambdadsinθ=mλ Third-order means m=3\displaystyle m = 3m=3, then 1.35×10−5sinθ=3(440×10−9)\displaystyle 1.35 \times 10^{-5}\sin \theta = 3(440 \times 10^{-9})1.35×10−5sinθ=3(440×10−9) This gives: sinθ=0.0977778\displaystyle \sin\theta = 0.0977778sinθ=0.0977778 θ=sin−10.0977778=5.61∘\displaystyle \theta = \sin^{-1} 0.0977778 = 5.61^{\circ}θ=sin−10.0977778=5.61∘
We use the diffraction maxima formula. dsinθ=mλ\displaystyle d\sin \theta = m\lambdadsinθ=mλ Third-order means m=3\displaystyle m = 3m=3, then 1.35×10−5sinθ=3(440×10−9)\displaystyle 1.35 \times 10^{-5}\sin \theta = 3(440 \times 10^{-9})1.35×10−5sinθ=3(440×10−9) This gives: sinθ=0.0977778\displaystyle \sin\theta = 0.0977778sinθ=0.0977778 θ=sin−10.0977778=5.61∘\displaystyle \theta = \sin^{-1} 0.0977778 = 5.61^{\circ}θ=sin−10.0977778=5.61∘