Given:
Focal length, \( f = 15 \, \text{cm} \)
Object distance, \( u = -30 \, \text{cm} \) (negative for real object)
Step 1: Mirror Formula The mirror formula is given by: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where: - \( f \) is the focal length, - \( v \) is the image distance, - \( u \) is the object distance.
Step 2: Solve for Image Distance \( v \) Substitute the given values into the mirror formula: \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{15} = \frac{1}{v} - \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{30} = \frac{2 + 1}{30} = \frac{3}{30} = \frac{1}{10} \] \[ v = 10 \, \text{cm} \]
Answer: The correct answer is option (c): 60 cm.
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?