The above 3 Figures (a) and (b) show refraction of a ray in air incident at `60^(@)` with the normal to a glass - air and water-air figures interface respectively. Predict the angle of refraction in glass when the angle of incidence in water is `45^(@)` with the normal to a water - glass interface [Fig. (c) ]

Explanation

1st Case :
Angle of incidence `i=60^(@)`
Angle of refraction `r=35^(@)`
`a_(mug)=(1)/(g_(mua))=(sin i)/(sin r)`
`a_(mug)=(sin 60^(@))/(sin 35^(@))=1.51`
2nd Case :
`a_(muw)=(sin 60^(@))/(sin 47^(@))=1.18`
`w_(mug)=a_(mug)xxw_(mua)`
`f=(a_(mug))/(a_(muw))=(1.51)/(1.18)=1.28`
3rd Case :
Angle of incidence `i=45^(@)`
Angle of refraction r = ?
`w_("rg")=(sin i)/(sin r)=(sin i)/(sin r)=1.28`
`Sin r=(sin 45^(@))/(1.28)=0.5525`
`sinr=sin33^(@)54'`
`r=33^(@)54'.`