Usual solution

In usual we asuume that $\sigma_{12}(k\sin\theta, \omega)=0, {\rm Im}\tilde{n}=0, \mu_1 = \mu_2 =1, \le... ...sin\theta, \omega) \right]_x = 0, \left[J_s (k\sin\theta, \omega) \right]_x = 0$, then for p-wave

$$(E_p - R_p ) \cos\theta = T_p \cos\theta''$$ (96)

$$n_1 (E_p + R_p ) = n_2 T_p$$ (97)

and for s-wave

$$E_s + R_s = T_s$$ (98)

$$n_1 \cos\theta (E_s + R_s ) = n_2 \cos\theta'' T_s$$ (99)

If we define

$$r_p \equiv \frac{R_p}{E_p}$$ (100)

$$t_p \equiv \frac{T_p}{E_p}$$ (101)

$$r_s \equiv \frac{R_s}{E_s}$$ (102)

$$t_s \equiv \frac{T_s}{E_s}$$ (103)

$$(1 - r_p ) \cos\theta = t_p \cos\theta''$$ (104)

$$n_1 ( 1 + r_p ) = n_2 t_p$$ (105)

and for s-wave

$$1 + r_s = t_s$$ (106)

$$n_1 \cos\theta (1 - r_s ) = n_2 \cos\theta'' t_s$$ (107)

And we finally get

$$r_p = \frac{-n_1 \cos\theta'' + n_2\cos\theta}{n_1 \cos\theta'' + n_2 \cos\theta }$$ (108)

$$t_p = \frac{2n_1 \cos\theta }{n_1 \cos\theta'' + n_2 \cos\theta}$$ (109)

$$r_s = \frac{n_1 \cos\theta - n_2 \cos\theta'' }{n_1 \cos\theta + n_2 \cos\theta'' }$$ (110)

$$t_s = \frac{2 n_1 \cos\theta}{n_1 \cos\theta + n_2 \cos\theta'' }$$ (111)

The reflectance $R$ is defined as the ratio of the reflected power (or flux) to the incident power

$$R = \frac{I' A \cos\theta'}{I A \cos\theta} = \frac{I'}{I} = \frac{n_1 R^2 /(2c\mu_0)}{ n_1 E^2 /(2c\mu_0)} = r^2$$ (112)

The radant flux density $I$ is given by the averaged Poynting vector $< {\bf E} \times {\bf H} > = \tilde{n} E^2/(2c\mu_0)$ In the same way The transmittance T may be

$$T = \frac{I'' A \cos\theta'' }{IA \cos\theta} = \frac{v_t \e... ... E^2 \cos\theta} = \frac{n_2 \cos\theta'' }{n_1 \cos\theta}t^2$$ (113)

図 4: He-Ne laser from air to water

図 5: He-Ne laser from water to air