The power \( P \) of a lens is related to its focal length \( f \) by the formula: \[ P = \frac{1}{f} \] where \( P \) is the power in diopters (D) and \( f \) is the focal length in meters. Given that the power of the corrective lens is \( P = -4.0 \, \text{D} \), we can find the focal length using the formula: \[ f = \frac{1}{P} \] Substituting \( P = -4.0 \, \text{D} \): \[ f = \frac{1}{-4.0} = -0.25 \, \text{m} = -25 \, \text{cm} \]
The correct option is (B) : concave lens of focal length -25 cm
The power of a lens (P) is related to its focal length (f) in meters by the formula:
\( P = \frac{100}{f(\text{in cm})} \Rightarrow f = \frac{100}{P} \)
Given: P = -4.0 D
\( f = \frac{100}{-4} = -25 \, \text{cm} \)
➤ A negative power indicates a concave lens.
✅ Correct answer: concave lens of focal length -25 cm
❌ Other options are incorrect because:
– Convex lenses have positive power.
– Focal length must match the power correctly.
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?