
Rules:
1)'D' is the distance b/w the virtual sources and the eyepiece, measured with a scale of least count 1mm.
Procedure: The virtual sources are at '0' on the optical bench (no zero error). The coordinate of the eyepiece is taken.
2)The fringe width and 'd1' and 'd2' (displacement method values) are measured with a micrometer screw of LC .01mm.
Procedure:: The cross-hair is set at a good- dark fringe (or at the image of a virtual source). Then it is moved to the 25th fringe (or to the image of the other virtual source). The difference between the 2 readings is taken.
Find the wavelength of the light used. What light is this?
wavelength=6.3x10^(-7)±4.8x10^(-9)
ReplyDeleteRed Light
[1.5636 +/- .468]*10^(-6)
ReplyDeleteinfrared lt
...ut infrared lt dikhegi kaise?..:p
i've assumed fringe separation given is for 25, not 1 fringe.
ReplyDeletepl confrm
ReplyDeleteshivanker, shaaan-how'd u get dat, i'm getting nowhere close..
ReplyDelete@anonymous: read the discussion here: http://anujkalia.blogspot.com/2011/03/fresnel-biprism.html
ReplyDeletehere are the various errors:
ReplyDeleteD : 2mm.
d: 1/2(.01+.01)mm.
fringe width: (1/25)*(.02)mm.
bhaiya
ReplyDeleteshudn't error of fringe width be (1/25)*(.01)mm?
bhaiya.. I always have this doubt..
ReplyDeleteAt some places Maximum Error is taken as the Least Count.. and at the other, twice of it..
which one should we prefer?
its a matter of 'fact' rather than 'preference'.
ReplyDeletewhen we take differences in readings of the instrument, error is 2*LC.
however, if we start from 0, and use an instrument without zero error, then error is 1*LC.
so what do we take as dx in the formulae for error analysis? :D
ReplyDelete