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

Light,whose electric field vectors are completely removed in the plane of incidence by using a good Polaroid,is incident on the surface of a prism at Brewster's angle. Choose the most suitable option for the phenomenon related to the prism.

The polarising angle for water is $53^\circ 4'$. If light is incident at this angle on the surface of water and reflected, the angle of refraction is:

$A$ ray of light is incident at a polarising angle such that its deviation is $24^{\circ}$. Then the angle of incidence is: (in $^{\circ}$)

$A$ solid glass sphere of refractive index $n=\sqrt{3}$ and radius $R$ contains a spherical air cavity of radius $\frac{R}{2}$,as shown in the figure. $A$ very thin glass layer is present at the point $O$ so that the air cavity (refractive index $n=1$) remains inside the glass sphere. An unpolarized,unidirectional and monochromatic light source $S$ emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point $O$ and is fully polarized,then the angle of incidence at the inner surface of the glass sphere is $\theta$. The value of $\sin \theta$ is

$A$ person lives in a high-rise building on the bank of a river $50\, m$ wide. Across the river is a well-lit tower of height $40\, m$. When the person,who is at a height of $10\, m$,looks through a polarizer at an appropriate angle at light of the tower reflecting from the river surface,he notes that the intensity of light coming from a distance $X$ from his building is the least,and this corresponds to the light coming from light bulbs at height $Y$ on the tower. The values of $X$ and $Y$ are respectively close to (refractive index of water $\simeq \frac{4}{3}$)