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In a YDSE with two identical slits, when...

In a YDSE with two identical slits, when the upper slit is covered with a thin, perfectly tranparent sheet of mica, the intensity at the centre of screen reduces ro `75%` of the initial value. The second minima is observed to above this point and third maxima below it. which of the following can not be a possible value of phase difference caused by the mica sheet

A

`(pi)/(3)`

B

`(7pi)/(2)`

C

`(10pi)/(3)`

D

`(11pi)/(3)`

Text Solution

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The correct Answer is:
A
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In a YDSE with two identical slits, when the upper slit is covered with a thin, perfectly tranparent sheet of mica, the intensity at the centre of screen reduces ro 75% of the initial value second minima is observed to the above this point and third maxima below it which of the following can not be a possible value of phase difference caused by the mica sheet

In a regular YDSE, when thin film of refractive index mu is placed in front of the upper slit then it is observed that the intensity at the central point becomes half of the original intensity. It is also observed that the initial 3^(rd) maxima is now below the central point and the initial 4^(th) minima is above the central point. Now, a film of refractive index mu_(1) and thickness same as the above film. is put in the front of the lower slit also. It is observed that whole fringe pattern shifts by one fringe width. What is the value of mu_(1) ?

In YDSE using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.5 and thickness 2 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the plane of slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the light.

In YDSE using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the plane of slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the light.

Young's double-slit experiment setup with ligth of wavelength lambda = 6000 Å , distance between two slit in 2 mm and distance between the plane of slits and the screen. Is 2 m. The slits are of equal intensity. When a sheet of glass of refractive index 1.5 (which permtis only a fraction eta of the incident light to pass through) and thickness 8000 Å is placed in front of the lower slit, it is observed that the intensity at a point P, 0.15 mm above the central maxima, does not change. The phase difference at point P without inserting the slab is

The young's double slit experiment is done in a medium of refractive index 4//3 .A light of 600nm wavelength is falling on the slits having 0.45 mm separation .The lower slit S_(2) is covered by a thin glass sheet of thickness 10.4mu m and refractive index 1.5 .the intereference pattern is observed ona screen placed 1.5m from the slits are shown (a) Find the location of the central maximum (bright fringe with zero path difference)on the y-axis. (b) Find the light intensity at point O relative to the maximum fringe intensity. (c )Now,if 600nm light is replaced by white light of range 400 to700 nm find the wavelength of the light that from maxima exactly point O .[All wavelengths in this problem are for the given medium of refractive index 4//3 .Ignore dispersion]

Minimum thickness of a mica sheet having mu=(3)/(2) which should be placed in front of one of the slits in YDSE is required to reduce the intensity at the centre of screen to half of maximum intensity is

A monochromatic beam of light of 6000 Å is used in YDSE set-up. The two slits are covered with two thin films of equal thickness t but of different refractive indices as shown in figure. Considering the intensity of the incident beam on the slits to be I_(0) , find the point on the screen at which intensity is I_(0) and is just above the central maxima. (Assume that there is no change in intensity of the light after passing through the films.) Consider t = 6 mu m, d = 1 mm, and D = 1 m, where d and D have their usual meaning. Give your answer in mm.

In a YDSE using monochromatic visible light, the distance between the plate of slits and the screen is 1.7 m. At point P on the screen which is directly in front of the upper slit, maximum path is observed. Now, the screen is moved 50 cm closer to the plane of slits. Point P now lies between third and fouth minima above the central maxima and the intensity at P in one-fourth of the maxima intensity on the screen. Find the value of n.

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