When a parallel beam of light enters a water surface obliquely, the beam undergoes refraction due to the change in the speed of light when passing from one medium (air) to another (water). Let's break down the effect of this refraction on the width of the beam:
When light passes from air (where the refractive index is approximately 1) into water (with a refractive index of about 1.33), the change in speed causes the light to bend according to Snell's law. The refractive index (\( n \)) is related to the angle of incidence (\( \theta_1 \)) and the angle of refraction (\( \theta_2 \)) as:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
where:When the light enters the water obliquely, the beam bends, causing the light rays to spread out more (or less) depending on the direction of the incident light. Since the speed of light in water is slower, the light rays are refracted towards the normal, causing the beam to become narrower in the direction of propagation.
When a parallel beam of light enters the water surface obliquely, the width of the beam decreases due to the refraction of light as it slows down and bends towards the normal. The greater the angle of incidence, the more pronounced the narrowing of the beam.
A current element X is connected across an AC source of emf \(V = V_0\ sin\ 2πνt\). It is found that the voltage leads the current in phase by \(\frac{π}{ 2}\) radian. If element X was replaced by element Y, the voltage lags behind the current in phase by \(\frac{π}{ 2}\) radian.
(I) Identify elements X and Y by drawing phasor diagrams.
(II) Obtain the condition of resonance when both elements X and Y are connected in series to the source and obtain expression for resonant frequency. What is the impedance value in this case?
In the given reaction sequence, the structure of Y would be: