Question

solve this if you can @irrelavent answers will be reported​

Answer 1

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\put(1,1.2){\line(1,0){6.5}}\end{picture}$

$\underline{\blacksquare\:\:\:\footnotesize{\red{Figure}}}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.3 mm}\put(0,0){\line(1,0){30}}\put(0,0){\line(0,1){20}}\put(0,20){\line(1,0){30}}\put(30,20){\line(0,-1){20}}\put(0,10){\line(-2,-3){10}}\linethickness{0.1 mm}\put(-10,10){\line(1,0){40}}\put(0,10){\line(5,3){17}}\put(17,20){\line(5,-3){13}}\put(17,20){\line(0,-1){20}}\put(-7,8){ \alpha_{max} }\put(14,16){ \theta_{c} }\footnotesize{\put(4,11){ (90\degree-\theta_{c}) }}\put(-2,20){A}\put(31,20){D}\put(-2,0){B}\put(31,0){C}\put(-5,13){ n_2 }\put(5,16){ n_1 }\put(17,20){\vector(1,0){7}}\end{picture}$

$\underline{\blacksquare\:\:\:\footnotesize{\red{SolutioN}}}$

$\footnotesize{\text{For the maximum value of }\alpha\:\text{, ray come out from CD }}$

$\footnotesize{\text{it should suffer limiting of internal reflection at AD }}$

$\footnotesize{\text{for the refraction at the surface AD}}$

$\footnotesize{n_1\sin\theta_c=n_2\sin90\degree}$

$\footnotesize{\sin\theta_c=\dfrac{n_2}{n_1}}$

$\footnotesize{\text{for the refraction at the surface AB}}$

$\footnotesize{n_2\sin\alpha_{max}=n_1\sin(90\degree-\theta_c)}$

$\footnotesize{\sin\alpha_{max}=\dfrac{n_1}{n_2}\cos\theta_c}$

$\footnotesize{\sin\alpha_{max}=\dfrac{n_1}{n_2}\cos[\sin^{-1}\dfrac{n_2}{n_1}]}$

$\boxed{\red{\footnotesize{\alpha_{max}=\sin^{-1}[\dfrac{n_1}{n_2}\cos\big(\sin^{-1}\dfrac{n_2}{n_1}\big)]}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\put(1,1.2){\line(1,0){6.5}}\end{picture}$

Answer 2

Answer:

ummm okay ..... take Care .....