Question-223847

Question Number 223847 by mr W last updated on 06/Aug/25

Commented by mr W last updated on 06/Aug/25

two mirrors form an angle α=8° as shown. a light ray starts from point A with an angle θ. such that the light ray returns to its origin after some reflections on the mirrors, find the minimum and the maximum value of θ.

$${two}\:{mirrors}\:{form}\:{an}\:{angle}\:\alpha=\mathrm{8}° \\ $$$${as}\:{shown}.\:{a}\:{light}\:{ray}\:{starts}\:{from}\: \\ $$$${point}\:{A}\:{with}\:{an}\:{angle}\:\theta.\:{such}\:{that} \\ $$$${the}\:{light}\:{ray}\:{returns}\:{to}\:{its}\:{origin} \\ $$$${after}\:{some}\:{reflections}\:{on}\:{the} \\ $$$${mirrors},\:{find}\:{the}\:{minimum}\:{and} \\ $$$${the}\:{maximum}\:{value}\:{of}\:\theta. \\ $$

Answered by mahdipoor last updated on 06/Aug/25

in A=P_1 ⇒ θ_1 and l_1 (=AO) in B=P_2 ⇒ θ_2 =θ_1 +α and l_2 =((sinθ_1 )/(sinθ_2 )).l_1 .... in P_(m+1) ⇒ θ_(m+1) =θ_1 +mα and l_(m+1) =((sinθ_1 )/(sin(θ_(m+1) ))).l_1 , if P_(m+1) =P_m ⇒ m=2n n≥1 , l_(m+1) =l_1 ⇒((sin(θ_1 ))/(sin(θ_(2n+1) )))=((sin(θ_1 ))/(sin(θ_1 +2nα)))=1 i} (θ_1 +2nα)−θ_1 =2kπ ⇒ α=(k/n)π=pπ p∈Q^+ for ∀θ_1 ii} (θ_1 +2nα)+θ_1 =(2k+1)π ⇒ θ_1 =kπ−nα+(π/2) ,,, α=(8/(360))×2π=(2/(45))π , case ii , 0<θ_1 <2π θ_1 =(k−((2n)/(45))+(1/2))π using AI (in renge : 1,100) min ⇒ π/90 = 2^° (m=23) max ⇒ 358π/90 ⇒ 358^° (m=3)

$${in}\:{A}={P}_{\mathrm{1}} \:\Rightarrow\:\theta_{\mathrm{1}} \:\:\:{and}\:\:{l}_{\mathrm{1}} \left(={AO}\right) \\ $$$${in}\:{B}={P}_{\mathrm{2}} \:\Rightarrow\:\theta_{\mathrm{2}} =\theta_{\mathrm{1}} +\alpha\:\:{and}\:\:{l}_{\mathrm{2}} =\frac{{sin}\theta_{\mathrm{1}} }{{sin}\theta_{\mathrm{2}} }.{l}_{\mathrm{1}} \\ $$$$…. \\ $$$${in}\:{P}_{{m}+\mathrm{1}} \:\Rightarrow\:\theta_{{m}+\mathrm{1}} =\theta_{\mathrm{1}} +{m}\alpha\:\:\:{and}\:\:\:{l}_{{m}+\mathrm{1}} =\frac{{sin}\theta_{\mathrm{1}} }{{sin}\left(\theta_{{m}+\mathrm{1}} \right)}.{l}_{\mathrm{1}} \\ $$$$, \\ $$$${if}\:{P}_{{m}+\mathrm{1}} ={P}_{{m}} \:\Rightarrow\:{m}=\mathrm{2}{n}\:\:{n}\geqslant\mathrm{1}\:\:,\:{l}_{{m}+\mathrm{1}} ={l}_{\mathrm{1}} \\ $$$$\Rightarrow\frac{{sin}\left(\theta_{\mathrm{1}} \right)}{{sin}\left(\theta_{\mathrm{2}{n}+\mathrm{1}} \right)}=\frac{{sin}\left(\theta_{\mathrm{1}} \right)}{{sin}\left(\theta_{\mathrm{1}} +\mathrm{2}{n}\alpha\right)}=\mathrm{1} \\ $$$$\left.\mathrm{i}\right\}\:\left(\theta_{\mathrm{1}} +\mathrm{2}{n}\alpha\right)−\theta_{\mathrm{1}} =\mathrm{2}{k}\pi\:\Rightarrow\: \\ $$$$\alpha=\frac{{k}}{{n}}\pi={p}\pi\:\:\:\:\:{p}\in\mathrm{Q}^{+} \:\:\:\:{for}\:\:\:\forall\theta_{\mathrm{1}} \\ $$$$\left.\mathrm{ii}\right\}\:\left(\theta_{\mathrm{1}} +\mathrm{2}{n}\alpha\right)+\theta_{\mathrm{1}} =\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\Rightarrow\: \\ $$$$\theta_{\mathrm{1}} ={k}\pi−{n}\alpha+\frac{\pi}{\mathrm{2}} \\ $$$$\:,,, \\ $$$$\alpha=\frac{\mathrm{8}}{\mathrm{360}}×\mathrm{2}\pi=\frac{\mathrm{2}}{\mathrm{45}}\pi\:,\:{case}\:\mathrm{ii}\:,\:\mathrm{0}<\theta_{\mathrm{1}} <\mathrm{2}\pi \\ $$$$\theta_{\mathrm{1}} =\left({k}−\frac{\mathrm{2}{n}}{\mathrm{45}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\pi \\ $$$${using}\:{AI}\:\:\left({in}\:{renge}\::\:\mathrm{1},\mathrm{100}\right) \\ $$$${min}\:\Rightarrow\:\pi/\mathrm{90}\:=\:\mathrm{2}^{°} \:\:\:\left({m}=\mathrm{23}\right) \\ $$$${max}\:\Rightarrow\:\mathrm{358}\pi/\mathrm{90}\:\Rightarrow\:\mathrm{358}^{°} \:\:\left({m}=\mathrm{3}\right) \\ $$

Answered by nikif99 last updated on 06/Aug/25

if 1 reflection→∡B=90°⇒θ=90−8=72 if 3 reflections (ABCBA)→∡C=90°⇒ ∡CBO=72⇒∡ABC=180−72−72=36⇒ θ=90−36=54 if 5 reflections (red lines)→∡D=90°⇒ ∡DCO=72=∡ACB, ∡BCD=36, DBC=54 ∡ABC=72⇒θ=36 similarly, if 7 reflections (red and blue lines) θ=18 when trying 9 reflections→ impossible (in “triangle” ABC, ∡ABC=144, ∡BCA=36, ∡BAC=0) min θ=18°, max θ=72°

$$ \\ $$$${if}\:\mathrm{1}\:{reflection}\rightarrow\measuredangle{B}=\mathrm{90}°\Rightarrow\theta=\mathrm{90}−\mathrm{8}=\mathrm{72} \\ $$$${if}\:\mathrm{3}\:{reflections}\:\left({ABCBA}\right)\rightarrow\measuredangle{C}=\mathrm{90}°\Rightarrow \\ $$$$\measuredangle{CBO}=\mathrm{72}\Rightarrow\measuredangle{ABC}=\mathrm{180}−\mathrm{72}−\mathrm{72}=\mathrm{36}\Rightarrow \\ $$$$\theta=\mathrm{90}−\mathrm{36}=\mathrm{54} \\ $$$${if}\:\mathrm{5}\:{reflections}\:\left({red}\:{lines}\right)\rightarrow\measuredangle{D}=\mathrm{90}°\Rightarrow \\ $$$$\measuredangle{DCO}=\mathrm{72}=\measuredangle{ACB},\:\measuredangle{BCD}=\mathrm{36},\:{DBC}=\mathrm{54} \\ $$$$\measuredangle{ABC}=\mathrm{72}\Rightarrow\theta=\mathrm{36} \\ $$$${similarly},\:{if}\:\mathrm{7}\:{reflections}\:\left({red}\:{and}\right. \\ $$$$\left.{blue}\:{lines}\right)\:\theta=\mathrm{18} \\ $$$${when}\:{trying}\:\mathrm{9}\:{reflections}\rightarrow\:{impossible} \\ $$$$\left({in}\:“{triangle}''\:{ABC},\:\measuredangle{ABC}=\mathrm{144},\right. \\ $$$$\left.\measuredangle{BCA}=\mathrm{36},\:\measuredangle{BAC}=\mathrm{0}\right) \\ $$$${min}\:\theta=\mathrm{18}°,\:{max}\:\theta=\mathrm{72}° \\ $$

Commented by nikif99 last updated on 06/Aug/25

Answered by mr W last updated on 07/Aug/25

Commented by mr W last updated on 07/Aug/25

α_1 =θ+α α_2 =α_1 +α=θ+2α ... α_n =θ+nα such that the light ray returns back to its origin at n^(th) reflection, α_n =90°. i.e. θ+nα=90° θ=90°−nα>0 ⇒n<((90)/8) ⇒n≤11 that means θ may have 11 possible values: θ_n =90°−n8° with 1≤n≤11 θ_(min) =90−11×8°=2° θ_(max) =90−1×8°=82°

$$\alpha_{\mathrm{1}} =\theta+\alpha \\ $$$$\alpha_{\mathrm{2}} =\alpha_{\mathrm{1}} +\alpha=\theta+\mathrm{2}\alpha \\ $$$$… \\ $$$$\alpha_{{n}} =\theta+{n}\alpha \\ $$$${such}\:{that}\:{the}\:{light}\:{ray}\:{returns}\:{back} \\ $$$${to}\:{its}\:{origin}\:{at}\:{n}^{{th}} \:{reflection},\:\alpha_{{n}} =\mathrm{90}°. \\ $$$${i}.{e}.\:\theta+{n}\alpha=\mathrm{90}° \\ $$$$\theta=\mathrm{90}°−{n}\alpha>\mathrm{0}\:\Rightarrow{n}<\frac{\mathrm{90}}{\mathrm{8}}\:\Rightarrow{n}\leqslant\mathrm{11} \\ $$$${that}\:{means}\:\theta\:{may}\:{have}\:\mathrm{11}\:{possible} \\ $$$${values}: \\ $$$$\theta_{{n}} =\mathrm{90}°−{n}\mathrm{8}°\:{with}\:\mathrm{1}\leqslant{n}\leqslant\mathrm{11} \\ $$$$\theta_{{min}} =\mathrm{90}−\mathrm{11}×\mathrm{8}°=\mathrm{2}° \\ $$$$\theta_{{max}} =\mathrm{90}−\mathrm{1}×\mathrm{8}°=\mathrm{82}° \\ $$

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