Let \( I_0 \) be the intensity of unpolarized light incident on the first polaroid.
- The transmitted light intensity after passing through the first polaroid is:
\(I_1 = \frac{I_0}{2}.\)
- The intensity after passing through the second polaroid at an angle \( \theta \) is:
\(I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta.\)
- Similarly, for the third polaroid rotated by \( 90^\circ - \theta \), the transmitted intensity is:
\(I_3 = I_2 \cos^2 (90^\circ - \theta) = \frac{I_0}{2} \cos^2 \theta \sin^2 \theta.\)
- The intensity will be maximum when \( \sin 2\theta = 1 \), which occurs at \( \theta = 45^\circ \).
The Correct answer is: \(45^\circ\)
Consider the sound wave travelling in ideal gases of $\mathrm{He}, \mathrm{CH}_{4}$, and $\mathrm{CO}_{2}$. All the gases have the same ratio $\frac{\mathrm{P}}{\rho}$, where P is the pressure and $\rho$ is the density. The ratio of the speed of sound through the gases $\mathrm{v}_{\mathrm{He}}: \mathrm{v}_{\mathrm{CH}_{4}}: \mathrm{v}_{\mathrm{CO}_{2}}$ is given by
Due to presence of an em-wave whose electric component is given by \( E = 100 \sin(\omega t - kx) \, NC^{-1} \), a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as:
Let \( \alpha, \beta \) be the roots of the equation \( x^2 - ax - b = 0 \) with \( \text{Im}(\alpha) < \text{Im}(\beta) \). Let \( P_n = \alpha^n - \beta^n \). If \[ P_3 = -5\sqrt{7}, \quad P_4 = -3\sqrt{7}, \quad P_5 = 11\sqrt{7}, \quad P_6 = 45\sqrt{7}, \] then \( |\alpha^4 + \beta^4| \) is equal to: