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: \[ \boxed{\text{it behaves like a concave lens of focal length 125 cm}}. \]
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]
\(XPQY\) is a vertical smooth long loop having a total resistance \(R\), where \(PX\) is parallel to \(QY\) and the separation between them is \(l\). A constant magnetic field \(B\) perpendicular to the plane of the loop exists in the entire space. A rod \(CD\) of length \(L\,(L>l)\) and mass \(m\) is made to slide down from rest under gravity as shown. The terminal speed acquired by the rod is _______ m/s. 