Calculating Electromagnetic Wave Speed in Liquid
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.
Determining Angles of Incidence and Refraction
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:
$\theta_2 = \theta_1 - \text{deviation} = 45^{\circ} - 15^{\circ} = 30^{\circ}$
Applying Snell's Law
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:
$1 \times \sin(45^{\circ}) = n_{liquid} \times \sin(30^{\circ})$
$n_{liquid} = \frac{\sin(45^{\circ})}{\sin(30^{\circ})} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Calculating Speed in Liquid
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 $2.1 \times 10^8 \text{ m/s}$.
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Consider following statements for refraction of light through prism, when angle of deviation is minimum.
A. The refracted ray inside prism becomes parallel to the base.
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E. Angle of refraction becomes double of prism angle. Choose the correct answer from the options given below:
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A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is :
