This problem involves calculating the speed of an electromagnetic wave as it passes from air into a liquid, using the principles of refraction and Snell's Law.
The wave normal makes an angle of $45^{\circ}$ with the vertical. Since the liquid surface is horizontal, the vertical direction is normal to the surface. Therefore, the angle of incidence ($\theta_1$) is:
$\theta_1 = 45^{\circ}$
The wave deviates by $15^{\circ}$. Deviation is the difference between the angle of incidence and the angle of refraction ($\theta_2$). Assuming the wave bends towards the normal (as it enters a potentially denser medium), the angle of refraction is:
$\theta_2 = \theta_1 - \text{deviation} = 45^{\circ} - 15^{\circ} = 30^{\circ}$
Snell's Law relates the angles of incidence and refraction to the refractive indices of the two media:
$n_{air} \sin(\theta_1) = n_{liquid} \sin(\theta_2)$
The refractive index of air ($n_{air}$) is approximately 1. Substituting the known values:
$1 \times \sin(45^{\circ}) = n_{liquid} \times \sin(30^{\circ})$
$n_{liquid} = \frac{\sin(45^{\circ})}{\sin(30^{\circ})} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$
The refractive index ($n$) of a medium is defined as the ratio of the speed of light in vacuum ($c$) to the speed of light in the medium ($v$): $n = c/v$. Since the speed of light in air ($v_{air}$) is given as $3 \times 10^8 \text{ m/s}$, and $n_{air} \approx 1$, we can consider $c \approx v_{air}$.
Therefore, the speed in the liquid ($v_{liquid}$) can be found using:
$n_{liquid} = \frac{v_{air}}{v_{liquid}}$
$v_{liquid} = \frac{v_{air}}{n_{liquid}}$
Substituting the values:
$v_{liquid} = \frac{3 \times 10^8 \text{ m/s}}{\sqrt{2}}$
$v_{liquid} = \frac{3}{\sqrt{2}} \times 10^8 \text{ m/s} \approx 2.1213 \times 10^8 \text{ m/s}$
Rounding to one decimal place, the speed of the electromagnetic wave in the liquid is approximately $2.1 \times 10^8 \text{ m/s}$.
Two points of monochromatic and coherent sources of light of wavelength $\lambda$ each, are placed as shown in figure. The initial phase difference between the sources is zero, ($D \gg d$). Mark the correct statement(s).