The image positions are calculated lens by lens using the lens formula:
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \implies v = \frac{uf}{u + f}.\]
For lens 1 (\(f_1 = 10 \, \text{cm}, u = -30 \, \text{cm}\)):
\[v = \frac{(-30) \times 10}{-30 + 10} = \frac{-300}{-20} = 15 \, \text{cm}.\]
The image forms 15 cm to the right of the first lens.
For lens 2 (\(f_2 = -10 \, \text{cm}, u = 10 \, \text{cm}\)):
\[v = \frac{10 \times (-10)}{10 - 10} = \infty.\]
The image is formed at infinity due to parallel rays after the second lens.
For lens 3 (\(f_3 = 30 \, \text{cm}, u = -\infty\)):
\[v = f_3 = 30 \, \text{cm}.\]
Thus, the final image is formed 30 cm to the right of the third lens.
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: