- #1
pardesi
- 339
- 0
Q.Suppose that the the light rays in young's Double Slit experiment fall at an angle [tex] \theta=\sin^{-1}(\frac{\lambda}{2d})[/tex] .Then prove that the the Point P which is the symmetric point on screen between two slits is the centre of central minimum
Necessary Formulaes
for minima the waves should interfere destructively so the path difference [tex]\Delta x=\frac{n\lambda}{2}[/tex]
ATTEMPT
i am not so sure of this but we can have a plane [tex]\pi,\pi^{'}[/tex] with inclination [tex]\theta[/tex] to the plane of slits and through the slits.the [tex]\perp[/tex] distance of the Point P from the plane determines it's phase so the phase difference is difference between the length of perpendiculars from P to the planes [tex]=d \sin \theta=\frac{\lambda}{2}[/tex] which produces destructive interference.
Doubt
My doubt lies on the fact that how do we apply difference in optical lengths concept here
Necessary Formulaes
for minima the waves should interfere destructively so the path difference [tex]\Delta x=\frac{n\lambda}{2}[/tex]
ATTEMPT
i am not so sure of this but we can have a plane [tex]\pi,\pi^{'}[/tex] with inclination [tex]\theta[/tex] to the plane of slits and through the slits.the [tex]\perp[/tex] distance of the Point P from the plane determines it's phase so the phase difference is difference between the length of perpendiculars from P to the planes [tex]=d \sin \theta=\frac{\lambda}{2}[/tex] which produces destructive interference.
Doubt
My doubt lies on the fact that how do we apply difference in optical lengths concept here