To determine the intensity of light after passing through two polaroids, we apply Malus's Law. Here's a step-by-step explanation:
Therefore, the intensity of the emergent light after passing through both polaroids is \(\frac{3I_0}{4}\).

The intensity of the emergent light,
\(I = I_0 × \frac{3}{4}\)
So, the correct option is (C): \( \frac{3I_0}{4}\)
| List-I | List-II | ||
| P | If \(n = 2\) and \(\alpha = 180°\), then all the possible values of \(\theta_0\) will be | I | \(30\degree\) or \(0\degree\) |
| Q | If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\theta_0\) will be | II | \(60\degree\) or \(0\degree\) |
| R | If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\phi_0\) will be | III | \(45\degree\) or \( 0\degree\) |
| S | If \(n = \sqrt2\) and \(\theta_0 = 45°\), then all the possible values of \(\alpha\) will be | IV | \(150\degree\) |
| \[0\degree\] | |||
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.
One of the major surfaces of Light Reaction is giving rise to organic energy molecules, such as ATP and NADPH, which are especially necessary for the subsequent occurrence of the dark reaction that works without the presence of sunlight.