The angle of incidence is found to be twice the angle of refraction when a ray of light passes from vacuum into a medium of refractive index $\mu$. The angle of incidence will be

If the wavelength of light in vacuum is $\lambda$,the wavelength in a medium of refractive index $n$ will be

Consider a light ray travelling in air incident into a medium of refractive index $\sqrt{2n}$. The incident angle is twice that of the refracting angle. Then,the angle of incidence will be

$A$ glass slab consists of thin uniform layers of progressively decreasing refractive indices $(RI)$ such that the $RI$ of any layer is $\mu - m \Delta \mu$. Here, $\mu$ and $\Delta \mu$ denote the $RI$ of the $0^{\text{th}}$ layer and the difference in $RI$ between any two consecutive layers, respectively. The integer $m = 0, 1, 2, 3, \ldots$ denotes the number of the successive layers. $A$ ray of light from the $0^{\text{th}}$ layer enters the $1^{\text{st}}$ layer at an angle of incidence of $30^{\circ}$. After undergoing the $m^{\text{th}}$ refraction, the ray emerges parallel to the interface. If $\mu = 1.5$ and $\Delta \mu = 0.015$, the value of $m$ is:

$A$ light ray is incident on a glass slab at an angle of incidence of $60^o$. If the angle between the reflected ray and the refracted ray is $90^o$,what is the refractive index of the glass slab?

An electromagnetic wave of wavelength $\lambda_0$ (in vacuum) passes from $P$ towards $Q$ crossing three different media of refractive index $\mu, 2\mu$ and $3\mu$ respectively as shown in the figure. Let $\phi_P$ and $\phi_Q$ be the phase of the wave at points $P$ and $Q$. Find the phase difference $\phi_Q - \phi_P$. [Take: $\mu = 1$]