A fish is at a real depth of 3.6 m in water (refractive index = 4/3). Calculate its apparent depth. If the angle of incidence in air is 42 degrees and the angle of refraction in water is 30 degrees, c

The real depth of the fish, $d_{real} = 3.6 \, \text{m}$.
The refractive index of water, $n = \frac{4}{3}$.

The relationship between real depth, apparent depth, and refractive index is given by:

The apparent depth of the fish is $\boxed{\text{2.7 m}}$.

c) i)

Angle of incidence in air, $i = 42^\circ$.
Angle of refraction in water, $r = 30^\circ$.
The refractive index of air, $n_{air} = 1$.

Snell's Law states $n_{air} \sin i = n_{water} \sin r$.

The refractive index of water is $\boxed{1.34}$ (rounded to two decimal places).

c) ii)

Angle of refraction in water, $r = 20^\circ$.
Angle of incidence in air, $i = \phi$.
Refractive index of air, $n_{air} = 1$.
Refractive index of water, $n_{water} = 1.3382$ (from part c i).

Angle $\phi$ is $\boxed{27.2^\circ}$ (rounded to one decimal place).

18. (a)
Two conditions for total internal reflection to occur are:

• Light must travel from an optically denser medium to an optically rarer medium.
• The angle of incidence in the denser medium must be greater than the critical angle for the interface between the two media.