This is the experimental setup that we'll be analyzing for error. (but lets do the theory first.)
1)Why are we using the bi-prism at all?? 2 marks
2)Why should the angle of the bi-prism be small?? 2 marks
3)Suggest a method for measuring the distance between the 2 virtual sources formed by the bi-prism. 4 marks.
In the diagram, I've also shown the view from the eyepiece. Also notice the red cross-hair. It can be moved by turning the screw of a micrometer, hence allowing us to measure the distance between the fringes!!
Hint: you'd like to use the eyepiece for Q3.
click the diagram to enlarge.
ReplyDelete1] to get 2 virtual sources above, below the actual one
ReplyDelete2]to find 'd'~ 2[mu-1][angle of prism][dist b/w source & prism]
here for small angle, we assumed
angle deviation~ [mu-1][angle of prism]
3]'d'~ 2[mu-1][angle of prism][dist b/w source & prism]
pls confirm
1+.5+1=2.5 marks..
ReplyDeletehow do you propose measuring the angle of the prism? your budget for new apparatus is Rs. 500.
1. to get 2 virtual "coherent" sources..(the slits can be sometimes tedious)
ReplyDelete2. for such small angle of prism, deviation produced for 'any' incident ray can be safely approximated to (angle of prism)(µ-1) and thus we obtain "2" sources who's apparent position doesn't change with the position of the screen
3. distance b/w virtual sources = 2(deviation)(dist b/w source & biprism)
or if we know the wavelength of light used.. it will be simply = (wavelength used)(distance b/w source & screen)/(fringe width)
or maybe i can measure the wavelength of light used by using a newton's color wheel(i know this one is quite crazy... :P )
1st is correct. 2nd needs to be corrected.
ReplyDeletewe don't know the wavelength of the light used..thats what needs finding.
hint: use a convex lens, which is readily available in the lab.
i've corrected the diagram. the double-slits are not used.
ReplyDelete2)small angle to minimise dispersion?
ReplyDeletewe can use the eyepiece to measure the size of the image of the source(point-sized) formed by a convex lens..
ReplyDeletethis is given by the formula 1.22(lambda)(focal length)/(aperture)
thus we can find lambda..
is that feasible?
this is not class 12 physics..this is proper JEE physics.
ReplyDeletewhat about displacement method?
ReplyDeletestill seems tough..
ReplyDeleteyou could see the 2 virtual sources thru the eyepiece..
ReplyDeleteyeah.. The same just struck me.. Was earlier focussing on finding wavelength..
ReplyDeleteWe can directly measure the separation of srcs in the 2 images in displacement method and thus the actual separation will be given by the gm of the 2..
Nice...
We cud not use the eyepiece directly to measure the separation b/w the sources because it shows us only the image "focussed" on it.. Just like a screen.. Am i right bhaiya?
1to make two coherent sources of same intensity.2to keep the distance betweem the two arbitary sources small.3.by using convex lens we can find the hieght of images formed by the sources simply by lens law
ReplyDelete2)'The distance has to be kept small to observe a clear interference pattern' should be included.
ReplyDelete3)The lens method (I hope you mean the lens formula) is very crude. Distances on the optical bench are measured with an error of +/- 1mm, while thru the eyepiece, they can be measured upto +/- .01 mm.