Kindly ans the following cases study in single thread

Solution,
17. Velocity of a wave (c) is defined as the product of its wavelength ($\lambda$) and frequency (f).
$c=\lambda f$
Velocity of a wave depends on the refractive index of the medium. It decreases with the increase in the refractive index of the material as:
$$\begin{gathered} c\text{'}=\frac{c}{n} \\ where, n is the refractive index of the medium \end{gathered}$$
Frequency is more fundamental and doesn't change with refractive index of the medium. Thus, the wavelength of a wave changes with the refractive index of the medium.
$$\begin{gathered} c\text{'}=f\lambda \text{'}=\frac{c}{n}=\frac{\lambda f}{n} \\ Thus, \\ \lambda \text{'}=\frac{\lambda }{n} \end{gathered}$$ 
Q17.
Given,
Wavelength of the monochromatic wave, $\lambda =589 nm$
refractive index of the water, $n_w=1.33=\frac{4}{3}$
Let the frequency of the wave be f.
In air (n=1), any electromagnetic wave travels with a speed of light, i.e., $c=3\times {10}^8 m/s$
Thus,
$$\begin{gathered} \lambda f=c \\ 589\times {10}^{-9}m\times f=3\times {10}^{8}m/s \\ f=\frac{3\times {10}^{8}m/s}{589\times {10}^{-9}m}=5.1\times {10}^{14}Hz \end{gathered}$$
Frequency of the wave will remain same.
wavelength of the wave in water,
$\lambda \text{'}=\frac{\lambda }{n}=\frac{589nm}{1.33}=442 nm$
Thus, ans is option (D)

Q18.
Speed of the refracted wave is:
$$\begin{gathered} c\text{'}=\frac{c}{n}=\frac{3\times {10}^8}{1.33}=2.25\times {10}^{8}m/s \\ Ans \left(A\right) \end{gathered}$$