In a biprism experiment, a steady interference pattern is observed on the screen kept at a distance of 100 cm using a light of wavelength $5000$ Å. Without changing the distance between the virtual images of the slit, the source of light is replaced by a source of wavelength $6400$ Å. Now, to reduce the fringe width by $20\%$ of its initial value, the screen should be moved

A ray of light is incident at polarising angle $\theta$ on air-glass interface. If $\lambda_a$ and $\lambda_g$ are the wavelengths of light in air and glass respectively then

In biprism experiment, the maximum intensity is $I_0$. If the path difference between the two interfering waves is $\dfrac{\lambda}{3}$, then intensity at the point on the screen is
[$\sin 30^\circ = \cos 60^\circ = 0.5$, $\sin 60^\circ = \cos 30^\circ = \sqrt{3}/2$]

In Young's double slit experiment, the wavelength of light used is $\lambda$. The intensity on the screen at a point for path difference '$\lambda$' is 'X'. The intensity at the point for path difference $\left(\dfrac{\lambda}{6}\right)$ is ($\cos 180^\circ = -1$, $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$)