In a typical young's double slit experiment a point source of monochromatic light is kept as shown in the figure. If the source is given an instantaneous velocity v=1 mm per second towards the screen, then the instantaneous velocity of central maxima is given as `alphaxx10^(-beta)cm//s` upward in scientific notation find the value of `alpha+beta`
_S01_033_Q01.png)
In a typical young's double slit experiment a point source of monochromatic light is kept as shown in the figure. If the source is given an instantaneous velocity v=1 mm per second towards the screen, then the instantaneous velocity of central maxima is given as `alphaxx10^(-beta)cm//s` upward in scientific notation find the value of `alpha+beta`
Text Solution
Verified by Experts
_S01_033_S01.png)
The central maxima `(dy)/(D)=sqrt(d^(2)+x^(2))-x=x[1+(d^(2))/(2x^(2))]-x=(d^(2))/(2x)`
`y=(Dd)/(2x)implies(dy)/(dt)=-(Dd)/(2x^(2))((dx)/(dt))=((1xx0.01)/(2xx0.5xx0.5))xx(0.001)=0.02mm//s`
`impliesy=2xx10^(-3)cm//simpliesalpha+beta=5`
|
Topper's Solved these Questions
WAVE OPTICS
ALLEN |Exercise Exercise 1 (Check your Grasp)|75 VideosView PlaylistWAVE OPTICS
ALLEN |Exercise Exercise 2 (Brain Teasures)|22 VideosView PlaylistWAVE OPTICS
ALLEN |Exercise Example 14|1 VideosView PlaylistUNIT & DIMENSIONS, BASIC MATHS AND VECTOR
ALLEN |Exercise Exercise (J-A)|7 VideosView Playlist
Similar Questions
Explore conceptually related problems
A vessel ABCD of 10 cm width has two small slits S_(1) and S_(2) sealed with idebtical glass plates of equal thickness. The distance between the slits is 0.8 mm . POQ is the line perpendicular to the plane AB and passing through O, the middle point of S_(1) and S_(2) . A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure. Calculate the position of the central bright fringe on the other wall CD with respect of the line OQ . Now, a liquid is poured into the vessel and filled up to OQ . The central bright fringe is fiund to be at Q. Calculate the refractive index of the liquid.
Watch solution
In an interference arrangement similar to young's double slit experiment the slits S_(1) and S_(2) are illuminated with coherent microwave sources each of frequency 10^(6)Hz the sources are synchronized to have zero phase difference. the slits are separated by a distance d=150.0m. The intensity I(theta) is measured as a funtion of theta , where theta is defined as shown. if I_(0) is the maximum intensity then I(theta) for 0lethetale90^(@) is given by:
Watch solution
While conduction the Young's double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S_(1),S_(2)) emitting light of wavelength 600nm. The student mistakenly placed the screen parallel to the x-z plane (for zgt0) at a distance D=3 m from the mid-point of S_(1) , S_(2) , as shown schematically in the figure. The distance between the sources d=0.6003 mm . The origin O is at the intersection of the screen and the line joining S_(1)S_(2) . Which of the following is (are) true of the intensity pattern of the screen?
Watch solution
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]
Watch solution
A Young's double slit interference arrangement with slits S_(1) and S_(2) is immersed in water (refractive index =4//3 ) as shown in the figure. The positions of maxima on the surface of water are given by x^(2) = p^(2)m^(2)lambda^(2)-d^(2) , where lambda is the wavelength of light in air (reflactive index = 1), 2d is the separation between the slits and m is an integer. The value of P is ..........
Watch solution
In the figure shown S is a point monochromatic light source of frequency 6xx10^(14)Hz .M is a concave mirror of radius of curvature 20 cm and L is a thin converging lens of focal length 3.75 cm AB is the principal axis of M and L. Light reflected from the mirror and refracted from the lens in succession reaches the screen. An interference pattern is obtained on the screen by this arrangement. Q. If the lens is replaced by another converging lens of focal length (10)/(3)cm and the lens is shifted towards right by 2.5 cm then
Watch solution
In figure S is a monochromatic point source emitting light of wavelength lambda=500 nm . A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L_(1) and L_(2) by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 mm . The distance along the axis from S to L_(1) and L_(2) is 0.15 m , while that from L_(1) and L_(2) to O is 1.30 m . The screen at O is normal to SO . (a) If the 3^(rd) intensity maximum occurs at point P on screen, find distance OP . (b) If the gap between L_(1) and L_(2) is reduced from its original value of 0.5 mm , will the distance OP increases, decreases or remain the same?
Watch solution
An electromagnetic wave can be represented by E = A sin (kx- omega t + phi) , where E is electric field associated with wave, According this equation, for any value of x, E remains sinusoidal for -oolt t lt oo . Obviously this corresponds to an idealised situation because radiation from ordinary sources consists of finite size wavetrains. In general, electric field remains sinusoidal only for times of order tau_(c) ' which is called coherence time. In simpler language it means that for times of order tau_(c)' a wave will have a definite phase. The finite value of coherence time could be due to many factors, for example if radiating atom undergoes collision with another atom then wave train undergoes an abrupt phase change or due to the fact that an atom responsible for emitting radiation has a finite life time in the energy level from which it drops to lower energy level, while radiating. Concept of coherence time can be easily understood using young's double slit experiment. Let interference patten is observed around point P at time t , due to superposition of waves emanting from S_(1) and S_(2) at times t =(r_(1))/(c) and (r_(2))/(c) respectively, where r_(1) and r_(2) are the distances S_(1) P & S_(2)P . Obviously if (r_(2)-r_(1))/(c) lt lt tau_(e),{"where" " "c = 3xx10^(8)m//s} then, wavetrain arriving at point P from S_(1) & S_(2) will have a definite phase relationship and an interference pattern of good contranst will be obtained. If coherence time is of order 10^(-10) second and screen is placed at a very large distance from slits in the given figure, then:-
Watch solution
An electromagnetic wave can be represented by E = A sin (kx- omega t + phi) , where E is electric field associated with wave, According this equation, for any value of x, E remains sinusoidal for -oolt t lt oo . Obviously this corresponds to an idealised situation because radiation from ordinary sources consists of finite size wavetrains. In general, electric field remains sinusoidal only for times of order tau_(c) ' which is called coherence time. In simpler language it means that for times of order tau_(c)' a wave will have a definite phase. The finite value of coherence time could be due to many factors, for example if radiating atom undergoes collision with another atom then wave train undergoes an abrupt phase change or due to the fact that an atom responsible for emitting radiation has a finite life time in the energy level from which it drops to lower energy level, while radiating. Concept of coherence time can be easily understood using young's double slit experiment. Let interference patten is observed around point P at time t , due to superposition of waves emanting from S_(1) and S_(2) at times t =(r_(1))/(c) and (r_(2))/(c) respectively, where r_(1) and r_(2) are the distances S_(1) P & S_(2)P . Obviously if (r_(2)-r_(1))/(c) lt lt tau_(e),{"where" " "c = 3xx10^(8)m//s} then, wavetrain arriving at point P from S_(1) & S_(2) will have a definite phase relationship and an interference pattern of good contranst will be obtained. If coherence time is of order 10^(-10) second and screen is placed at a very large distance from slits in the given figure, then:-
Watch solution
In a YDSE experiment two slits S_(1) and S_(2) have separation of d=2mm the distance of the screen is D=(8)/(5) m source S starts moving from a very large distance towards S_(2) perpendicular to S_(1)S_(2) as shown in figure the wavelength of monochromatic light is 500 n. The number of maximas observed on the screen at point P as the moves towards S_(2) is
Watch solution