高考 physics 学习笔记 - Optics
光速： 真空中 c = 3 × 10 8 m/s c = 3 \times 10^8\,\text{m/s} c = 3 × 1 0 8 m/s
影的形成： 光沿直线传播遇到不透明物体，在物体后面形成阴影区域。
全反射： 光从光密介质射向光疏介质时，当入射角大于临界角时，光线全部被反射回光密介质。
临界角公式：sin C = 1 n \sin C = \frac{1}{n} sin C = n 1 n n n
反射定律： 反射角等于入射角，反射光线、入射光线和法线在同一平面内。
平面镜成像特点：
折射定律（Snell 定律）： n 1 sin θ 1 = n 2 sin θ 2 n_1 \sin\theta_1 = n_2 \sin\theta_2 n 1 sin θ 1 = n 2 sin θ 2
折射率： n = c v n = \frac{c}{v} n = v c v v v
全反射条件：
光从光密介质射向光疏介质 入射角大于或等于临界角 双缝干涉： 明纹位置 x = k ⋅ L λ d x = k \cdot \frac{L\lambda}{d} x = k ⋅ d L λ k = 0 , ± 1 , ± 2 , … k = 0, \pm 1, \pm 2, \ldots k = 0 , ± 1 , ± 2 , …
薄膜干涉： 光在薄膜前后两个表面反射的光叠加产生干涉。
单缝衍射： 中央明纹最宽最亮，两侧明纹逐渐变暗变窄。
爱因斯坦光电效应方程： E k = h ν − W 0 E_k = h\nu - W_0 E k = h ν − W 0
其中 h ν h\nu h ν W 0 W_0 W 0 E k E_k E k
截止频率： ν 0 = W 0 h \nu_0 = \frac{W_0}{h} ν 0 = h W 0
题目： 一束光从空气射入水中，入射角为 45 ° 45° 45° n = 4 3 n = \frac{4}{3} n = 3 4
解答：
步骤1：由折射定律 n 1 sin θ 1 = n 2 sin θ 2 n_1 \sin\theta_1 = n_2 \sin\theta_2 n 1 sin θ 1 = n 2 sin θ 2
步骤2：代入数据（空气 n 1 ≈ 1 n_1 \approx 1 n 1 ≈ 1 sin 45 ° = 4 3 sin θ 2 \sin 45° = \frac{4}{3} \sin\theta_2 sin 45° = 3 4 sin θ 2 2 2 = 4 3 sin θ 2 \frac{\sqrt{2}}{2} = \frac{4}{3} \sin\theta_2 2 2 = 3 4 sin θ 2
步骤3：解得： sin θ 2 = 3 2 8 \sin\theta_2 = \frac{3\sqrt{2}}{8} sin θ 2 = 8 3 2 θ 2 = arcsin 3 2 8 \theta_2 = \arcsin\frac{3\sqrt{2}}{8} θ 2 = arcsin 8 3 2
答案： 折射角为 arcsin 3 2 8 \arcsin\frac{3\sqrt{2}}{8} arcsin 8 3 2
题目： 光从玻璃（n = 1.5 n = 1.5 n = 1.5
解答：
步骤1：由全反射临界角公式 sin C = 1 n \sin C = \frac{1}{n} sin C = n 1
步骤2：代入数据： sin C = 1 1.5 = 2 3 \sin C = \frac{1}{1.5} = \frac{2}{3} sin C = 1.5 1 = 3 2
步骤3：计算临界角： C = arcsin 2 3 ≈ 41.8 ° C = \arcsin\frac{2}{3} \approx 41.8° C = arcsin 3 2 ≈ 41.8°
答案： 临界角约为 41.8 ° 41.8° 41.8°
题目： 用频率为 ν = 6 × 10 14 Hz \nu = 6 \times 10^{14}\,\text{Hz} ν = 6 × 1 0 14 Hz W 0 = 2.0 eV W_0 = 2.0\,\text{eV} W 0 = 2.0 eV h = 6.63 × 10 − 34 J⋅s h = 6.63 \times 10^{-34}\,\text{J·s} h = 6.63 × 1 0 − 34 J⋅s 1 eV = 1.6 × 10 − 19 J 1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J} 1 eV = 1.6 × 1 0 − 19 J
解答：
步骤1：计算光子能量： E = h ν = 6.63 × 10 − 34 × 6 × 10 14 = 3.98 × 10 − 19 J E = h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 3.98 \times 10^{-19}\,\text{J} E = h ν = 6.63 × 1 0 − 34 × 6 × 1 0 14 = 3.98 × 1 0 − 19 J
步骤2：换算为电子伏特： E = 3.98 × 10 − 19 1.6 × 10 − 19 ≈ 2.49 eV E = \frac{3.98 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.49\,\text{eV} E = 1.6 × 1 0 − 19 3.98 × 1 0 − 19 ≈ 2.49 eV
步骤3：由光电效应方程： E k = E − W 0 = 2.49 − 2.0 = 0.49 eV E_k = E - W_0 = 2.49 - 2.0 = 0.49\,\text{eV} E k = E − W 0 = 2.49 − 2.0 = 0.49 eV
答案： 光电子的最大初动能约为 0.49 eV 0.49\,\text{eV} 0.49 eV
光路可逆原理在折射和反射问题中经常用到 全反射问题必须先判断光是否从光密介质射向光疏介质 干涉问题注意区分明纹和暗纹的条件 光电效应问题中，光强影响光电流大小，频率影响光电子的最大初动能 一束光从水（n = 4 3 n = \frac{4}{3} n = 3 4 30 ° 30° 30° 求光从玻璃（n = 1.5 n = 1.5 n = 1.5 n = 4 3 n = \frac{4}{3} n = 3 4 用波长为 400 nm 400\,\text{nm} 400 nm 2.5 eV 2.5\,\text{eV} 2.5 eV 题目： 在双缝干涉实验中，双缝间距为 0.2 mm 0.2\,\text{mm} 0.2 mm 1 m 1\,\text{m} 1 m 600 nm 600\,\text{nm} 600 nm
解答：
步骤1：由双缝干涉公式，相邻明纹间距： Δ x = L λ d \Delta x = \frac{L\lambda}{d} Δ x = d L λ
步骤2：代入数据： Δ x = 1 × 600 × 10 − 9 0.2 × 10 − 3 = 6 × 10 − 7 2 × 10 − 4 = 3 × 10 − 3 m = 3 mm \Delta x = \frac{1 \times 600 \times 10^{-9}}{0.2 \times 10^{-3}} = \frac{6 \times 10^{-7}}{2 \times 10^{-4}} = 3 \times 10^{-3}\,\text{m} = 3\,\text{mm} Δ x = 0.2 × 1 0 − 3 1 × 600 × 1 0 − 9 = 2 × 1 0 − 4 6 × 1 0 − 7 = 3 × 1 0 − 3 m = 3 mm
答案： 相邻明纹间距为 3 mm 3\,\text{mm} 3 mm
题目： 一束光从空气射入玻璃砖（n = 1.5 n = 1.5 n = 1.5 45 ° 45° 45° 10 cm 10\,\text{cm} 10 cm
解答：
步骤1：由折射定律求折射角： sin 45 ° = 1.5 sin r ⟹ sin r = 2 / 2 1.5 = 2 3 \sin 45° = 1.5 \sin r \implies \sin r = \frac{\sqrt{2}/2}{1.5} = \frac{\sqrt{2}}{3} sin 45° = 1.5 sin r ⟹ sin r = 1.5 2 /2 = 3 2 r ≈ 28.1 ° r \approx 28.1° r ≈ 28.1°
步骤2：在玻璃砖内的水平位移： x = d tan r = 10 × tan 28.1 ° ≈ 10 × 0.534 = 5.34 cm x = d \tan r = 10 \times \tan 28.1° \approx 10 \times 0.534 = 5.34\,\text{cm} x = d tan r = 10 × tan 28.1° ≈ 10 × 0.534 = 5.34 cm
步骤3：侧移量（入射光线延长线与出射光线的垂直距离）： δ = d sin ( i − r ) / cos r = 10 × sin ( 45 ° − 28.1 ° ) / cos 28.1 ° \delta = d \sin(i - r) / \cos r = 10 \times \sin(45° - 28.1°) / \cos 28.1° δ = d sin ( i − r ) / cos r = 10 × sin ( 45° − 28.1° ) / cos 28.1° = 10 × sin 16.9 ° / cos 28.1 ° ≈ 10 × 0.291 / 0.882 = 3.3 cm = 10 \times \sin 16.9° / \cos 28.1° \approx 10 \times 0.291 / 0.882 = 3.3\,\text{cm} = 10 × sin 16.9°/ cos 28.1° ≈ 10 × 0.291/0.882 = 3.3 cm
答案： 侧移量约为 3.3 cm 3.3\,\text{cm} 3.3 cm
题目： 用频率为 ν 1 \nu_1 ν 1 E k 1 E_{k1} E k 1 ν 2 \nu_2 ν 2 E k 2 E_{k2} E k 2
解答：
步骤1：由光电效应方程： E k 1 = h ν 1 − W 0 E_{k1} = h\nu_1 - W_0 E k 1 = h ν 1 − W 0 E k 2 = h ν 2 − W 0 E_{k2} = h\nu_2 - W_0 E k 2 = h ν 2 − W 0
步骤2：两式相减： E k 1 − E k 2 = h ( ν 1 − ν 2 ) E_{k1} - E_{k2} = h(\nu_1 - \nu_2) E k 1 − E k 2 = h ( ν 1 − ν 2 )
步骤3：解得普朗克常量： h = E k 1 − E k 2 ν 1 − ν 2 h = \frac{E_{k1} - E_{k2}}{\nu_1 - \nu_2} h = ν 1 − ν 2 E k 1 − E k 2
步骤4：代入第一个方程求逸出功： W 0 = h ν 1 − E k 1 = ( E k 1 − E k 2 ) ν 1 ν 1 − ν 2 − E k 1 W_0 = h\nu_1 - E_{k1} = \frac{(E_{k1} - E_{k2})\nu_1}{\nu_1 - \nu_2} - E_{k1} W 0 = h ν 1 − E k 1 = ν 1 − ν 2 ( E k 1 − E k 2 ) ν 1 − E k 1 = E k 1 ν 1 − E k 2 ν 1 − E k 1 ν 1 + E k 1 ν 2 ν 1 − ν 2 = E k 1 ν 2 − E k 2 ν 1 ν 1 − ν 2 = \frac{E_{k1}\nu_1 - E_{k2}\nu_1 - E_{k1}\nu_1 + E_{k1}\nu_2}{\nu_1 - \nu_2} = \frac{E_{k1}\nu_2 - E_{k2}\nu_1}{\nu_1 - \nu_2} = ν 1 − ν 2 E k 1 ν 1 − E k 2 ν 1 − E k 1 ν 1 + E k 1 ν 2 = ν 1 − ν 2 E k 1 ν 2 − E k 2 ν 1
答案： h = E k 1 − E k 2 ν 1 − ν 2 h = \dfrac{E_{k1} - E_{k2}}{\nu_1 - \nu_2} h = ν 1 − ν 2 E k 1 − E k 2 W 0 = E k 1 ν 2 − E k 2 ν 1 ν 1 − ν 2 W_0 = \dfrac{E_{k1}\nu_2 - E_{k2}\nu_1}{\nu_1 - \nu_2} W 0 = ν 1 − ν 2 E k 1 ν 2 − E k 2 ν 1
折射问题： 先画光路图，确定入射角和折射角，再用折射定律全反射问题： 先判断是否从光密介质射向光疏介质，再计算临界角干涉问题： 区分双缝干涉（Δ x = L λ / d \Delta x = L\lambda/d Δ x = L λ / d 光电效应： 理解截止频率、最大初动能与频率的线性关系在几何光学中，光路是可逆的。如果光线从A到B经过某光学系统，那么从B发出的光线将沿原路返回A。这个原理在解题中非常有用，可以简化复杂光路的分析。
题目： 一束光以入射角 i = 60 ° i = 60° i = 60° A = 60 ° A = 60° A = 60° n = 3 n = \sqrt{3} n = 3
解答：
步骤1：在第一个界面，由折射定律： sin 60 ° = 3 sin r 1 \sin 60° = \sqrt{3} \sin r_1 sin 60° = 3 sin r 1 3 2 = 3 sin r 1 ⟹ sin r 1 = 1 2 ⟹ r 1 = 30 ° \frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1 \implies \sin r_1 = \frac{1}{2} \implies r_1 = 30° 2 3 = 3 sin r 1 ⟹ sin r 1 = 2 1 ⟹ r 1 = 30°
步骤2：由棱镜的几何关系，r 1 + r 2 = A r_1 + r_2 = A r 1 + r 2 = A r 2 = A − r 1 = 60 ° − 30 ° = 30 ° r_2 = A - r_1 = 60° - 30° = 30° r 2 = A − r 1 = 60° − 30° = 30°
步骤3：在第二个界面，由折射定律： 3 sin r 2 = sin e \sqrt{3} \sin r_2 = \sin e 3 sin r 2 = sin e 3 sin 30 ° = sin e \sqrt{3} \sin 30° = \sin e 3 sin 30° = sin e 3 2 = sin e ⟹ e = 60 ° \frac{\sqrt{3}}{2} = \sin e \implies e = 60° 2 3 = sin e ⟹ e = 60°
步骤4：偏向角： δ = i + e − A = 60 ° + 60 ° − 60 ° = 60 ° \delta = i + e - A = 60° + 60° - 60° = 60° δ = i + e − A = 60° + 60° − 60° = 60°
答案： 光线的偏向角为 60 ° 60° 60°
考试技巧： 三棱镜问题中，光线通过棱镜时向底面偏折。最小偏向角条件为 r 1 = r 2 = A / 2 r_1 = r_2 = A/2 r 1 = r 2 = A /2 i = e i = e i = e
题目： 用两种频率分别为 ν 1 \nu_1 ν 1 ν 2 \nu_2 ν 2 ν 1 > ν 2 \nu_1 > \nu_2 ν 1 > ν 2 U 1 U_1 U 1 U 2 U_2 U 2
解答：
步骤1：由光电效应方程和遏止电压的关系： e U 1 = h ν 1 − W 0 eU_1 = h\nu_1 - W_0 e U 1 = h ν 1 − W 0 e U 2 = h ν 2 − W 0 eU_2 = h\nu_2 - W_0 e U 2 = h ν 2 − W 0
步骤2：两式相减： e ( U 1 − U 2 ) = h ( ν 1 − ν 2 ) e(U_1 - U_2) = h(\nu_1 - \nu_2) e ( U 1 − U 2 ) = h ( ν 1 − ν 2 )
步骤3：解得普朗克常量： h = e ( U 1 − U 2 ) ν 1 − ν 2 h = \frac{e(U_1 - U_2)}{\nu_1 - \nu_2} h = ν 1 − ν 2 e ( U 1 − U 2 )
步骤4：代入第一个方程求逸出功： W 0 = h ν 1 − e U 1 = e ( U 1 − U 2 ) ν 1 ν 1 − ν 2 − e U 1 W_0 = h\nu_1 - eU_1 = \frac{e(U_1 - U_2)\nu_1}{\nu_1 - \nu_2} - eU_1 W 0 = h ν 1 − e U 1 = ν 1 − ν 2 e ( U 1 − U 2 ) ν 1 − e U 1 = e ν 1 U 1 − e ν 1 U 2 − e U 1 ν 1 + e U 1 ν 2 ν 1 − ν 2 = e ( U 1 ν 2 − U 2 ν 1 ) ν 1 − ν 2 = \frac{e\nu_1 U_1 - e\nu_1 U_2 - eU_1 \nu_1 + eU_1 \nu_2}{\nu_1 - \nu_2} = \frac{e(U_1 \nu_2 - U_2 \nu_1)}{\nu_1 - \nu_2} = ν 1 − ν 2 e ν 1 U 1 − e ν 1 U 2 − e U 1 ν 1 + e U 1 ν 2 = ν 1 − ν 2 e ( U 1 ν 2 − U 2 ν 1 )
答案： h = e ( U 1 − U 2 ) ν 1 − ν 2 h = \dfrac{e(U_1 - U_2)}{\nu_1 - \nu_2} h = ν 1 − ν 2 e ( U 1 − U 2 ) W 0 = e ( U 1 ν 2 − U 2 ν 1 ) ν 1 − ν 2 W_0 = \dfrac{e(U_1 \nu_2 - U_2 \nu_1)}{\nu_1 - \nu_2} W 0 = ν 1 − ν 2 e ( U 1 ν 2 − U 2 ν 1 )
常见错误： 遏止电压对应的是光电子的最大初动能，即 e U c = E k eU_c = E_k e U c = E k
题目： 在双缝干涉实验中，用波长为 λ = 500 nm \lambda = 500\,\text{nm} λ = 500 nm Δ x = 2 mm \Delta x = 2\,\text{mm} Δ x = 2 mm λ ′ = 700 nm \lambda' = 700\,\text{nm} λ ′ = 700 nm
解答：
步骤1：由双缝干涉公式 Δ x = L λ d \Delta x = \dfrac{L\lambda}{d} Δ x = d L λ Δ x ∝ λ \Delta x \propto \lambda Δ x ∝ λ
步骤2：因此： Δ x ′ Δ x = λ ′ λ \frac{\Delta x'}{\Delta x} = \frac{\lambda'}{\lambda} Δ x Δ x ′ = λ λ ′
步骤3：代入数据： Δ x ′ = Δ x ⋅ λ ′ λ = 2 × 700 500 = 2.8 mm \Delta x' = \Delta x \cdot \frac{\lambda'}{\lambda} = 2 \times \frac{700}{500} = 2.8\,\text{mm} Δ x ′ = Δ x ⋅ λ λ ′ = 2 × 500 700 = 2.8 mm
答案： 相邻明纹间距变为 2.8 mm 2.8\,\text{mm} 2.8 mm
考试技巧： 双缝干涉中，波长越长，条纹间距越大。红光条纹间距大于绿光，绿光大于蓝光。条纹间距与波长成正比。
题目： 光纤由折射率为 n 1 = 1.5 n_1 = 1.5 n 1 = 1.5 n 2 = 1.3 n_2 = 1.3 n 2 = 1.3
解答：
步骤1：全反射条件：光从光密介质射向光疏介质，入射角大于临界角
步骤2：临界角 C C C sin C = n 2 n 1 = 1.3 1.5 = 13 15 \sin C = \dfrac{n_2}{n_1} = \dfrac{1.3}{1.5} = \dfrac{13}{15} sin C = n 1 n 2 = 1.5 1.3 = 15 13
步骤3：故 C = arcsin 13 15 ≈ 60.1 ° C = \arcsin\dfrac{13}{15} \approx 60.1° C = arcsin 15 13 ≈ 60.1°
步骤4：光在纤芯中传播时，入射角 θ \theta θ θ > C ≈ 60.1 ° \theta > C \approx 60.1° θ > C ≈ 60.1°
答案： 入射角必须大于 arcsin 13 15 ≈ 60.1 ° \arcsin\dfrac{13}{15} \approx 60.1° arcsin 15 13 ≈ 60.1°
考试技巧： 光纤通信利用全反射原理，光在纤芯中不断全反射向前传播。数值孔径 N A = n 1 2 − n 2 2 NA = \sqrt{n_1^2 - n_2^2} N A = n 1 2 − n 2 2
题目： 用频率为 ν = 6 × 10 14 Hz \nu = 6 \times 10^{14}\,\text{Hz} ν = 6 × 1 0 14 Hz U c = 0.5 V U_c = 0.5\,\text{V} U c = 0.5 V e = 1.6 × 10 − 19 C e = 1.6 \times 10^{-19}\,\text{C} e = 1.6 × 1 0 − 19 C h = 6.63 × 10 − 34 J⋅s h = 6.63 \times 10^{-34}\,\text{J·s} h = 6.63 × 1 0 − 34 J⋅s
解答：
步骤1：由光电效应方程：E k = h ν − W 0 E_k = h\nu - W_0 E k = h ν − W 0
步骤2：遏止电压与最大初动能的关系：e U c = E k eU_c = E_k e U c = E k
步骤3：E k = 1.6 × 10 − 19 × 0.5 = 8 × 10 − 20 J E_k = 1.6 \times 10^{-19} \times 0.5 = 8 \times 10^{-20}\,\text{J} E k = 1.6 × 1 0 − 19 × 0.5 = 8 × 1 0 − 20 J
步骤4：W 0 = h ν − E k = 6.63 × 10 − 34 × 6 × 10 14 − 8 × 10 − 20 = 3.98 × 10 − 19 − 8 × 10 − 20 = 3.18 × 10 − 19 J W_0 = h\nu - E_k = 6.63 \times 10^{-34} \times 6 \times 10^{14} - 8 \times 10^{-20} = 3.98 \times 10^{-19} - 8 \times 10^{-20} = 3.18 \times 10^{-19}\,\text{J} W 0 = h ν − E k = 6.63 × 1 0 − 34 × 6 × 1 0 14 − 8 × 1 0 − 20 = 3.98 × 1 0 − 19 − 8 × 1 0 − 20 = 3.18 × 1 0 − 19 J
步骤5：W 0 = 3.18 × 10 − 19 1.6 × 10 − 19 ≈ 1.99 eV W_0 = \dfrac{3.18 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.99\,\text{eV} W 0 = 1.6 × 1 0 − 19 3.18 × 1 0 − 19 ≈ 1.99 eV
步骤6：截止频率 ν 0 = W 0 h = 3.18 × 10 − 19 6.63 × 10 − 34 ≈ 4.8 × 10 14 Hz \nu_0 = \dfrac{W_0}{h} = \dfrac{3.18 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.8 \times 10^{14}\,\text{Hz} ν 0 = h W 0 = 6.63 × 1 0 − 34 3.18 × 1 0 − 19 ≈ 4.8 × 1 0 14 Hz
答案： 逸出功约为 1.99 eV 1.99\,\text{eV} 1.99 eV 4.8 × 10 14 Hz 4.8 \times 10^{14}\,\text{Hz} 4.8 × 1 0 14 Hz
常见错误： 遏止电压对应的是光电子的最大初动能，不是光子能量。e U c = E k = h ν − W 0 eU_c = E_k = h\nu - W_0 e U c = E k = h ν − W 0
题目： 自然光通过两个偏振片后，光强变为原来的 1 8 \dfrac{1}{8} 8 1 90 ° 90° 90°
解答：
步骤1：自然光通过第一个偏振片后，光强变为 I 0 2 \dfrac{I_0}{2} 2 I 0
步骤2：通过第二个偏振片后，由马吕斯定律：I = I 0 2 cos 2 θ I = \dfrac{I_0}{2}\cos^2\theta I = 2 I 0 cos 2 θ
步骤3：初始时 I = I 0 8 I = \dfrac{I_0}{8} I = 8 I 0 I 0 2 cos 2 θ = I 0 8 \dfrac{I_0}{2}\cos^2\theta = \dfrac{I_0}{8} 2 I 0 cos 2 θ = 8 I 0 cos 2 θ = 1 4 \cos^2\theta = \dfrac{1}{4} cos 2 θ = 4 1 cos θ = 1 2 \cos\theta = \dfrac{1}{2} cos θ = 2 1 θ = 60 ° \theta = 60° θ = 60°
步骤4：旋转 90 ° 90° 90° 60 ° + 90 ° = 150 ° 60° + 90° = 150° 60° + 90° = 150° ∣ 60 ° − 90 ° ∣ = 30 ° |60° - 90°| = 30° ∣60° − 90°∣ = 30°
步骤5：若夹角为 30 ° 30° 30° I ′ = I 0 2 cos 2 30 ° = I 0 2 × 3 4 = 3 I 0 8 I' = \dfrac{I_0}{2}\cos^2 30° = \dfrac{I_0}{2} \times \dfrac{3}{4} = \dfrac{3I_0}{8} I ′ = 2 I 0 cos 2 30° = 2 I 0 × 4 3 = 8 3 I 0
步骤6：若夹角为 150 ° 150° 150° I ′ = I 0 2 cos 2 150 ° = I 0 2 × 3 4 = 3 I 0 8 I' = \dfrac{I_0}{2}\cos^2 150° = \dfrac{I_0}{2} \times \dfrac{3}{4} = \dfrac{3I_0}{8} I ′ = 2 I 0 cos 2 150° = 2 I 0 × 4 3 = 8 3 I 0
答案： 旋转后的光强为 3 I 0 8 \dfrac{3I_0}{8} 8 3 I 0
考试技巧： 马吕斯定律：I = I 0 cos 2 θ I = I_0 \cos^2\theta I = I 0 cos 2 θ θ \theta θ
A[Optics] --> B[Key Concepts]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
D --> G[Real-world usage]
Optics is like understanding how mirrors and lenses bend light. Reflection is like a ball bouncing off a wall: the angle of incidence equals the angle of refraction. This is why you can see yourself in a mirror: light from your face bounces off the mirror and into your eyes.
Refraction is like light changing lanes on a highway. When light moves from air to water, it slows down and bends toward the normal. This is why a straw in water looks bent: the light from the submerged part changes direction as it exits the water. Understanding these bends lets you predict how lenses focus light and how mirrors redirect it.
Confusing the mirror formula with the lens formula. For mirrors, 1/f = 1/v + 1/u. For lenses, 1/f = 1/v - 1/u. Students often use the mirror formula for lenses or vice versa, leading to incorrect sign conventions and wrong focal length calculations.
Forgetting sign conventions in geometric optics. The sign convention determines whether distances and image sizes are positive or negative. For mirrors, distances in front are positive. For lenses, distances on the opposite side of the incoming light are positive. Mixing conventions gives wrong answers.
Confusing total internal reflection conditions. Total internal reflection occurs when light travels from a denser medium to a rarer medium AND the angle of incidence exceeds the critical angle. Students often forget one of these conditions, incorrectly predicting total internal reflection when it does not occur.
Mechanics - Wave mechanics and oscillatory motion underlying the wave theory of lightElectricity - Electromagnetic theory connecting electric and magnetic fields to light propagationAlgebra - Algebraic manipulation of lens and mirror formulas for image calculations