Question #102883
For obtaining Newton’s rings due to reflected light, a convex lens is placed on a plane glass plate and irradiated by monochromatic light. Obtain an expression showing the relation between radii of successive dark rings related to each other?
1
Expert's answer
2020-02-14T09:31:10-0500

As per the newton's ring experiment,




The ray of light which are getting refract form the convex lens and then getting incident on the plane mirror which is inclined at the 4545^\circ , after the reflection form the plane mirror, ray of light is getting incident on the lens which is place on the flat glass plate.

On the first surface of the plano-convex lens, ray of light does not getting deflect from the first surface of the lens,



ray of light will get deflect, when it will incident on the second surface,

So, the path difference(PD) =2μtcosr±λ22\mu t \cos r\pm \dfrac{\lambda}{2}

So, r =0,

PD=2t±λ2PD=2t\pm \dfrac{\lambda}{2}

at the center t=0

PD=λ2\dfrac{\lambda}{2}

For the Bright frindge,

PD=nλPD=n\lambda

2t±λ2=nλ2t\pm \dfrac{\lambda}{2}=n\lambda

2t=nλ±λ22t=n\lambda\pm\dfrac{\lambda}{2}

For the dark frindge,

PD=(n+12)λPD=(n+\dfrac{1}{2}) \lambda

2t±λ2=(n+12)λ2t\pm \dfrac{\lambda}{2}=(n+\dfrac{1}{2})\lambda

2t=nλ2t=n\lambda

So, radius of nth dark fringe

rn2=nRλr_n^2=nR\lambda

where r= radius of the fringe

n is the order of the fringe

R= Radius of the curvature of the lens

λ=\lambda= wavelength of the light



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