1.

In young's experiment, the upper slit is covered by a thin glass plate of refractive index 4/3 and of thickness 9lamda, where lamda is the wavelength of light used in the experiment. The lower slit is also covered by another glass plate of thickness 2D & refractive index 3/2, as shown in figure. If I_(0) is the intensity at point P due to slits S_(1) & S_(2) each, then:

Answer»

<P>Intensity at point `P` is `4I_(0)`
Two fringes have been shifted in upward direction after inserted of both the glass plates together.
Optical path difference between te waves from `S_(1)` & `S_(2)` at point `P` is `2lamda`
If the source `S` is shifted upwards by a SMALL distance `d_(2)` then the fringe ORIGINALLY at `P` after inserting the plates, shifts downwards by `D((d_(2))/(d_(1)))`

Solution :Optical path difference due to
`S_(1)` is `=t(mu_(1)-1)=9lamda(4/3-1)=3lamda`
Optical path DIFFER due to `S_(2)` is `=t(mu_(2)-1)=2lamda(3/2-1)=LAMDA`
Path diff. at point `PP=3lamda-lamda=2lamda`
Phase diff. `(2pi)/(lamda)xx2lamda=4pi`
So net intensity at point `P` is
`I_("net")=I_(0)+I_(0)+2I_(0)cos 4 pi`
`I_("net")=4I_(0)`
Optical path is consrved by `S_(1)` is more so fringe will shift at upper side.


Discussion

No Comment Found

Related InterviewSolutions