To solve this question, we need to determine the correct optical aid for a person suffering from both Myopia (nearsightedness) and Hypermetropia (farsightedness).
1. Understanding the Conditions:
- Myopia: Corrected using concave lenses.
- Hypermetropia: Corrected using convex lenses.
2. When Both Conditions Are Present:
A person with both conditions needs correction for both near and distant vision. This is achieved using a bi-focal lens, which has two parts: - The upper part corrects distant vision (myopia). - The lower part corrects near vision (hypermetropia).
3. Analyzing the Options:
- Bi-convex lens: Only for hypermetropia.
- Bi-concave lens: Only for myopia.
- Bi-focal lens: Corrects both.
- Bi-focal mirror: Not used for vision correction.
Final Answer:
Option (C) Bi-focal lens is correct.
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?