Answer to Question #100432 in Optics for OLAMIDE OJ

Question

A rectangular glass block (n=1.52) is placed inside a clean water contained in a basin. A layer of oil (n=1.44) floats on the water surface. A ray of light from air strikes the oil surface at an angle of incidence of 28° . Calculate the angle of refraction at the water-glass interface. (refractive index of air =1.00)

Answer 1

air-oil

$\frac{sin(\alpha_1)}{sin(\alpha_2)}=\frac{n_o}{n_a}$

oil-water

$\frac{sin(\alpha_2)}{sin(\alpha_3)}=\frac{n_w}{n_o}$

water-glass

$\frac{sin(\alpha_3)}{sin(\alpha_4)}=\frac{n_g}{n_w}$

combine this equation

$$sin(\alpha_4)=\frac{n_w}{n_g}\cdot \frac{n_o}{n_w}\cdot \frac{n_a}{n_o}\cdot sin(\alpha_1)$$ $$\alpha_4=arcsin(\frac{n_a}{n_g}\cdot sin(\alpha_1))$$

calculating

$\alpha_4=arcsin(\frac{1}{1.52}\cdot sin(28))=18^o$