When parallel rays from a distant source fall on a spherical glass ball, they refract at the surface, and the rays converge to form an image. Since the ball is a sphere, the incident rays are refracted at both the entry and exit points, forming a real image on the other side of the ball. To find the position of the image, we can apply the formula for the refraction at a spherical surface: \[ \frac{1}{f} = \left( \frac{n - 1}{R} \right) \]
where:
\( f \) is the focal length of the spherical ball,
\( n = 1.5 \) is the refractive index of the glass,
\( R = 15 \, \text{cm} \) is the radius of the spherical ball.
Substitute the values: \[ \frac{1}{f} = \frac{1.5 - 1}{15} = \frac{0.5}{15} = \frac{1}{30} \] So, the focal length \( f = 30 \, \text{cm} \). The ray diagram for this setup is shown below:

The image is formed at a distance of 30 cm from the center of the ball. Therefore, the final image is formed 30 cm away from the center on the opposite side of the incident light. Since the source is far away, the rays converge at this point after refracting through the spherical ball.
Calculate the angle of minimum deviation of an equilateral prism. The refractive index of the prism is \(\sqrt{3}\). Calculate the angle of incidence for this case of minimum deviation also.
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?