Total Power of Lenses in Contact:
When lenses are kept in contact, the effective power \( P_{eq} \) is the sum of the individual powers of each lens:
\[ P_{eq} = \sum P_i \]
Given that there are 5 identical lenses and the total power is 25 D, we have:
\[ 5P = 25 \implies P = \frac{25}{5} = 5 \, \text{D} \]
where \( P \) is the power of each individual lens.
Calculate the Focal Length:
The focal length \( f \) of a lens is related to its power \( P \) by:
\[ P = \frac{1}{f} \]
where \( f \) is in meters if \( P \) is in diopters (D).
Therefore:
\[ f = \frac{1}{P} = \frac{1}{5} = 0.2 \, \text{m} = 20 \, \text{cm} \]
Conclusion:
The focal length of each convex lens is 20 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?
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: