Question

(a)and(b)show the refraction of a ray in air incident at 60° with the normal to a glass-air and water-air 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
 

Answer 1

As per the given figure, for the glass-air interface:
The angle of incidence,i=60^º
The angle of refraction,r=35^º
The relative refractive index of glass with respect to air is given by Snell's law as:
μ_aw=\(\frac{sin\,i}{sin\,r}\)=\(\frac{sin60}{sin47}=\frac{0.8660}{0.7314}\)=1.184…………….(2)
Using(1)and(2), the relative refractive index of glass with respect to water can be obtained as:μwg/μaw=\(\frac{1.51}{1.184}\)=1.275
The following figure shows the situation involving the glass-water interface.
Angle of the interface,i=45º
Angle of refraction=r
From Snell's law,r can be calculated as:\(\frac{sin\,i}{sin\,r}\)=μ_wg=\(\frac{sin\,45^{\circ}}{sin\,r}\)=1.275=sinr=\(\frac{1}{\sqrt{1.274}}\)=0.5546
∴r=sin^-1(0.5546)=38.68º
Hence, the angle of refraction at the water-glass interface is 38.68º.