At what angle should a ray of light be incident on the face of a prism of refracting angle `60^(@)` so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Explanation

Critical angle :`i_(c)=sin^(-1)((1)/(n))`
`i_(c)=sin^(-1)((1)/(1.5))=sin^(-1)(0.6667)`
`=41^(@)49.`
Angle of incidence onn the second face:
`r_(2)=41^(@)49.`
Angle of refraction at the first face:
`r_(1)A-r_(2)`
`=60^(@)-41^(@)40.=18^(@)11`
Applying Snell.s law to the first face:
`n=(sin i)/(sin r_(1))`
[OR using `n_(1)` sin i =`n_(2)` sin `r_(1),n_(1)=1,n_(2)`=1.5]
`implies` sin i=n sin `r_(1)`
`implies` sin i=1.5 in `18^(@)11`
1.5(0.3121)=0.4682
ANgle of incidence on the first face:
i=`sin^(1)(0.482)=27^(@)55.`

(Award full marks for the answwers rounded off to 2 significant figures :1/1.5=0.67 , `i_(c)=42^(@),i=28^(@))`
Detailed Answer:
Refractive index =`mu` =1.5
Critical angke =`theta_(C)`
`therefore sin theta_(C)=(1)/(mu)`
or `si theta_(C)=(2)/(3)`
`therefore theta_(C)=42^(@)`
`therefore angle ADP=90^(@)-42^(@)=48^(@)`
`angle BAC=60^(@)`

`therefore angle APD=180^(@)-60^(@)-48^(@)=72^(@)`
`therefore angler =90^(@)-72^(@)=18^(@)`
`(sini)/(sinr)=mu`
Or, `(sini)/(sin18^(@))=1.5`
OR, sin i=`1.5xxsin 18^(@)`
OR, `sini=1.5xx0.31`
Or,`sini=0.465`
`therefore i=27.7^(@)`
The angl of incidence should be `27.7^(@)`