114年:視覺光學

陳先生右眼所需眼鏡的兩個稜鏡量分別為,2.83Δ基底在135度和5.00Δ基底37度,組合他的兩個稜鏡量為一,則最終稜鏡處方下列何者正確?(sin45=0.707; sin37=0.6; 5=2.236; 29=5.385; 37=6.083; 61=7.810)

A2.2Δ
B5.4Δ
C6.1Δ
D7.8Δ

詳細解析

本題觀念:

本題考稜鏡量的向量合成(prism compounding)。將兩個斜向稜鏡處方合併為單一稜鏡,需先將各稜鏡分解為水平(X軸)和垂直(Y軸)分量,再用畢氏定理(Pythagorean theorem)求合力大小。

題目提供的輔助數值:sin45°=0.707\sin 45° = 0.707sin37°=0.6\sin 37° = 0.6cos37°=0.8\cos 37° = 0.8(由 cos37°=1sin237°=0.8\cos 37° = \sqrt{1-\sin^2 37°} = 0.8 推導);5=2.236\sqrt{5} = 2.23629=5.385\sqrt{29} = 5.38537=6.083\sqrt{37} = 6.08361=7.810\sqrt{61} = 7.810

選項分析

計算過程(向量分解)

稜鏡1:2.83Δ,基底在135° P1x=2.83×cos135°=2.83×(0.707)=2.00ΔP_{1x} = 2.83 \times \cos 135° = 2.83 \times (-0.707) = -2.00\,\Delta P1y=2.83×sin135°=2.83×(+0.707)=+2.00ΔP_{1y} = 2.83 \times \sin 135° = 2.83 \times (+0.707) = +2.00\,\Delta

(注意:2.83222.83 \approx 2\sqrt{2},故 2.83×0.7072.002.83 \times 0.707 \approx 2.00,計算整齊)

稜鏡2:5.00Δ,基底在37°

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