The power of a thin convex lens placed in air is \( +4D \). The refractive index of the material of the convex lens is \( \frac{3}{2} \). If this convex lens is immersed in a liquid of refractive index \( \frac{5}{3} \), then:
It behaves like a convex lens of focal length 75 cm
It behaves like a convex lens of focal length 125 cm
It behaves like a concave lens of focal length 125 cm
It behaves like a concave lens of focal length 75 cm
Step 1: Using Lens Maker's Formula
The power of a lens in air is given by the lens maker's formula: \[ P_{\text{air}} = \left( \frac{n_L}{n_A} - 1 \right) \frac{100}{f} \] where \( P_{\text{air}} = +4D \), \( n_L = \frac{3}{2} \) (refractive index of lens), and \( n_A = 1 \) (refractive index of air). The focal length in air is: \[ f_{\text{air}} = \frac{100}{P_{\text{air}}} = \frac{100}{4} = 25 \text{ cm}. \]
Step 2: Finding Power in the Liquid Medium
When the lens is placed in a medium of refractive index \( n_M = \frac{5}{3} \), the power is given by: \[ P_{\text{liquid}} = \left( \frac{n_L}{n_M} - 1 \right) \frac{100}{f_{\text{air}}}. \] Substituting the values: \[ P_{\text{liquid}} = \left( \frac{\frac{3}{2}}{\frac{5}{3}} - 1 \right) \frac{100}{25}. \] Simplifying: \[ P_{\text{liquid}} = \left( \frac{3}{2} \times \frac{3}{5} - 1 \right) \times 4. \] \[ P_{\text{liquid}} = \left( \frac{9}{10} - 1 \right) \times 4. \] \[ P_{\text{liquid}} = \left( -\frac{1}{10} \right) \times 4. \] \[ P_{\text{liquid}} = -0.4D. \] Since the power is negative, the lens behaves as a concave lens. The focal length is: \[ f_{\text{liquid}} = \frac{100}{|P_{\text{liquid}}|} = \frac{100}{0.8} = 125 \text{ cm}. \]
Step 3: Conclusion
Thus, the lens behaves like a concave lens with a focal length of 125 cm: \[ \boxed{\text{it behaves like a concave lens of focal length 125 cm}}. \]
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?
Match the following: