Step 1: Use the mirror formula
The mirror formula relates the object distance \( u \), the image distance \( v \), and the focal length \( f \): \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where:
- \( f \) is the focal length,
- \( v \) is the image distance,
- \( u \) is the object distance.
Step 2: Substitute the given values
Given:
- Focal length \( f = -15 \, \text{cm} \) (for concave mirror, focal length is negative),
- Object distance \( u = -10 \, \text{cm} \) (object is always placed on the same side as the incoming light).
Substitute these values into the formula: \[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-10} \] \[ \frac{1}{v} = \frac{1}{-15} + \frac{1}{10} \] \[ \frac{1}{v} = \frac{-2 + 3}{30} \] \[ \frac{1}{v} = \frac{1}{30} \] \[ v = 30 \, \text{cm} \]
Answer:
Therefore, the image distance is \( 30 \, \text{cm} \). So, the correct answer is option (1).
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?