The direction of propagation of light is perpendicular to the wave front and is symmetric about the \( x \), \( y \), and \( z \) axes.
The angle made by the direction of wave propagation with the \( x \)-axis is the same as that with the \( y \)-axis and the \( z \)-axis.
Thus, the equation can be written as: \[ \cos \theta = \cos \beta = \cos \gamma \quad (\text{where } \alpha, \beta, \gamma \text{ are the angles made by light with the } x, y, z \text{ axes respectively}) \] Also, we know that \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \). Since the angles are equal, we have: \[ \cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \quad \Rightarrow \quad 3 \cos^2 \alpha = 1 \quad \Rightarrow \quad \cos \alpha = \frac{1}{\sqrt{3}} \] Thus, the angle is \( \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \).
Match List-I with List-II on the basis of two simple harmonic signals of the same frequency and various phase differences interacting with each other:
LIST-I (Lissajous Figure) | LIST-II (Phase Difference) | ||
---|---|---|---|
A. | Right handed elliptically polarized vibrations | I. | Phase difference = \( \frac{\pi}{4} \) |
B. | Left handed elliptically polarized vibrations | II. | Phase difference = \( \frac{3\pi}{4} \) |
C. | Circularly polarized vibrations | III. | No phase difference |
D. | Linearly polarized vibrations | IV. | Phase difference = \( \frac{\pi}{2} \) |
Choose the correct answer from the options given below:
The motion of an airplane is represented by the velocity-time graph as shown below. The distance covered by the airplane in the first 30.5 seconds is km.
The least acidic compound, among the following is
Choose the correct set of reagents for the following conversion: