The phase difference between the incident wave and the reflected wave is $180^o$ when a light ray:

  • A
    Enters into glass from air
  • B
    Enters into air from glass
  • C
    Enters into glass from diamond
  • D
    Enters into water from glass

Explore More

Similar Questions

In the figure below,$PQRS$ denotes the path followed by a ray of light as it travels through three media in succession. The absolute refractive indices of the media are $\mu_1, \mu_2$ and $\mu_3$,respectively. (The line segment $RS$ in the figure is parallel to $PQ$). Then,

$A$ 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:

The optical path of a monochromatic light is the same if it goes through $4.0 \, cm$ of glass or $4.5 \, cm$ of water. If the refractive index of glass is $1.53$,the refractive index of water is:

$A$ circular disc of radius $R$ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius $a$ (See figure). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with a transparent liquid of refractive index $\mu$ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

Difficult
View Solution

The angles of incidence and refraction of a monochromatic ray of light of wavelength $\lambda$ at an air-glass interface are $i$ and $r$,respectively. $A$ parallel beam of light with a small spread $\delta \lambda$ in wavelength about a mean wavelength $\lambda$ is refracted at the same air-glass interface. The refractive index $\mu$ of glass depends on the wavelength $\lambda$ as $\mu(\lambda)=a+b / \lambda^2$,where $a$ and $b$ are constants. Then,the angular spread in the angle of refraction of the beam is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo