Consider the figure given below. A ray of light is incident on one of the horizontal surfaces of glass slab at an angle of `60^(@)` and it just grazes along the adjacent vertical sides after refraction. Calculate the critical angle .

Explanation

For refraction at AB,
`(sin 60^(@))/(sinr)=mu`
`rArr" "sinr=(sin60^(@))/(mu)`
Also, from figure, `r=90^(@)-i_(c )`
`therefore" "sin(90^(@)-i_(c ))=(sin60^(@))/(mu)`
`rArr" "cosi_(c )=(sin60^(@))/(mu)"or "mu=(sin60^(@))/(cosi_(c ))`
For refraction at BC,
`sini_(c )=(1)/(mu)=(cosi_(c ))/(sin 60^(@))`
`rArr" "(sini_(c ))/(cosi_(c ))=(1)/(sin60^(@))or tan i_(c )=(2)/(sqrt3)`
`rArr" "i_(c )=tan^(-1)((2)/(sqrt3))=49.1^(@)`