Wave optics practice · Medium

The wavelength of the light used in Young’s double slit experiment is $\lambda$. The intensity at a point on the screen is $I$, where the path difference is $\frac{\lambda }{6}$. If ${{I_0}}$ denotes the maximum intensity, then the ratio of ${I\;and\;{I_0}}$ is

In Young’s double slit experiment the wavelength of light was changed from $7000\;\mathop {\mathrm{A}\;to\;}\limits^ \circ \;3500\;\mathop \mathrm{A}\limits^ \circ$. While doubling the separation between the slits which of the following is not true for this experiment

The two coherent sources of intensity that ratio $2: 8$ produce an interference pattern. The values of maximum and minimum intensities will be respectively. Where $I_1$ is the intensity of first source

For Young's double slit experiment, two statements are given below:

Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains constant.

Statement II : If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.

In the light of the above statements, choose the correct answer from the options given below:

In a single slit diffraction pattern with slit width $a$ and wavelength of light $\lambda$, find the angular position of first minima if screen distance is $D(D \gg a)$

An incident plane wavefront makes angle $53^{\circ}$ with surface of medium in air. The refracted wavefront makes angle $37^{\circ}$ with the surface. Refractive index of medium is :-

Astronomical wavelength increase due to doppler effect known as

Assertion: In interference, only redistribution of light energy occurs in the form of maximum and minimum.

Reason: Average energy in the interference pattern is smae as it would be if there were no interference.

What is the path difference of destructive interference

In young's experiment light of wavelength $4800 ~\mathop {\text{A}}\limits^ \circ$ is used. If one slit is covered with glass of refractive index $1.4$ and another slit is covered with a film of same thickness having refractive index $1.7,$ the centre fringe shifts to the position of $5^{\text {th}}$ bright fringe then find the thickness of film.

The principle of superposition is basic to the phenomenon of

In the phenomenon of diffraction of light, when blue light is used in the experiment instead of red light, then

In Young’s double slit experiment, the distance between the two slits is made half, then the fringe width will become

Assertion: If the phase difference between the light waves emerging from the slits of the Young's experiment is $\pi$-radian, the central fringe will be dark.

Reason: Phase difference is equal to $(2 \pi / \lambda)$ times the path difference.

In Young’s double slit experiment we get 60 fringes in the field of view of monochromatic light of wavelength ${4000\;\mathop \mathrm{A}\limits^ \circ }$. If we use monochromatic light of wavelength ${6000\;\mathop \mathrm{A}\limits^ \circ }$, then the number of fringes obtained in the same field of view are

In a Young's double slit experiment the angular width of a fringe formed on a distant screen is $1^{\circ}$. The wavelength of the light used is $6280\mathop {\text{A}}\limits^ \circ$. What is the distance between the two coherent sources?

The sky appears blue, because

In Young's double slit experiment if the separation between coherent sources is doubled and distance of screen from slit plane is halved, then fringe width becomes

${n^{th}}$ Bright fringe if red light $\left( {{\lambda _1} = 7500\mathop \mathrm{A}\limits^ \circ } \right)$ coincides with ${\left( {n + 1} \right)^{th}}$ bright fringe of green light $\left( {{\lambda _2} = 6000\mathop \mathrm{A}\limits^ \circ } \right)$. The value of $n = ?$

Wavefront of a wave has direction with wave motion

An optically active compound

${V_O}\;and\;{V_E}$ represent the velocities, ${\mu _O}\;and\;{\mu _E}$ the refractive indices of ordinary and extraordinary rays for a doubly refracting crystal. Then

In Young’s double slit experiment, distance between two slits is $0.28\;\mathrm{mm}$ and distance between slits and screen is $1.4\;\mathrm{m}$. Distance between central bright fringe and third bright fringe is $0.9\;\mathrm{cm}$, what is the wavelength of light used?

Plane polarised light is passed through a polaroid. On viewing through the polaroid we find that when the polariod is given one complete rotation about the direction of the light, one of the following is observed

In single slit diffraction pattern

Assertion : Polaroid glasses reduce glare from road or water etc.

Reason : Reflected waves are partially polarised.

A ray of light falls on a transparent glass slab with refractive index (relative to air) of $\mathrm{1}\mathrm{.62}$. The angle of incidence for which the reflected and refracted rays are mutually perpendicular is

When both the components of electric field of light waves are present such that one is stronger than the other and such light is viewed through a rotating analyser, one sees a maximum and a minimum of intensity but not complete darkness. This kind of light is called

Wavefront means

In Young’s double slit experiment, when violet light of wavelength $4358\mathop \mathrm{A}\limits^ \circ$ is used, the 84 fringe are seen in the field of view, but when sodium light of certain wavelength is used, then 62 fringes are seen in the field of view, the wavelength of sodium light is