Question

A ray of light incident on a glass slab of thickness t
and refractive index 1.5 at an angle of 60°. The time
taken by ray to cross the slab is (given c is speed of
light in air)
​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\implies\textsf{ A ray of light incidents on a glass slab at \sf 60^{\circ} } .\\\implies\textsf{ The refractive index of glass is 1.5 }\\\implies\textsf{ Speed of light is \bf{ c}.}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\implies\textsf{ The time taken to cross the glass slab .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

The thickness of the glass slab is t and the ray incidents at 60° . And the speed of the light is c that is 3 × 10⁸ m/s . So the time taken will be equal to Thickness of glass slab / Speed of light .

$\underline{\underline{\blue{\sf Diagram :- }}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){5}}\put(5,0){\line(-1,2){2}}\put(3,5){\line(0,-1){2}}\put(3,5){\vector(0,-1){0.7}}\multiput(3,0)(0,0.4){8}{\line(0,1){0.2}}\put(2.5,5){\vector( 1,-2){2}}\put(-0.3,2){\bf t }\put(0.5,1){ \bf \mu = 1.5 } \put(2.8, - 0.4){ \bf B } \put(5, - 0.4){ \bf C }\put(2.5, 4.1){ \bf A }\put(3.2, 2){ \bf 60^{\circ} }\qbezier(3,3)(3.3, 2.6)(3.5,3)\end{picture}$

For Figure may refer to attachment . Here we will produce A to B. So , In ∆ ABC ,

$\underline{\boldsymbol{\purple{ In \ \triangle ABC \ we \ have :- }}}$

$\sf:\implies\pink{ cos\ 60^{\circ} = \dfrac{AB}{AC}}\\\\\sf:\implies \dfrac{1}{2}= \dfrac{ t}{AC} \\\\\sf:\implies\boxed{\pink{\mathfrak{ AC = 2t }}}$

$\rule{200}2$

$\underline{\boldsymbol{\purple{ According\ to \ formula \ of \ time :- }}}$

$\sf:\implies \pink{ Time = \dfrac{Distance}{Speed}}\\\\\sf:\implies t = \dfrac{ AC}{v } \\\\\sf:\implies t = \dfrac{ 2t}{\dfrac{c}{\mu}}\qquad\qquad\bigg\lgroup \red{\bf v= \dfrac{c}{\mu}}\bigg\rgroup \\\\\sf:\implies t = \dfrac{ 1.5 \times 2 t}{c}\\\\\sf:\implies \boxed{\pink{\mathfrak{ Time = \dfrac{3t}{c}}}}$

$\underline{\blue{\sf \therefore Hence \ the \ time \ taken \ to \ cross \ is \ \bf{\dfrac{3t}{c} }. }}$