Question

A Newton's ring arrangement is
used with a source emitting two
wavelengths 6000 and 4500
angstrom. It is found that nth
dark ring due to lambda 1
coincide with n+1 th dark ring
of lambda 2, if radius of
curvature is 90 cm, find the
diameter of nth dark ring​

Answer 1

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HERE IS YOUR ANSWER

λ1=6000A0=6000×10−8cm for D2n

λ1=6000A0=6000×10−8cm for

D2nλ2=4500A0=4500×10−8cm for D2n+1

λ1=6000A0=6000×10−8cm for D2nλ2=4500A0=4500×10−8cm for D2n+1R=90cm

λ1=6000A0=6000×10−8cm for D2nλ2=4500A0=4500×10−8cm for D2n+1R=90cmμ=1

λ1=6000A0=6000×10−8cm for D2nλ2=4500A0=4500×10−8cm for D2n+1R=90cmμ=1Find rn=?

λ1=6000A0=6000×10−8cm for D2nλ2=4500A0=4500×10−8cm for D2n+1R=90cmμ=1Find rn=?D2nD2nD2n+1=4Rnλ=4Rnλ1=4R(n+1)λ2

λ1=6000A0=6000×10−8cm for D2nλ2=4500A0=4500×10−8cm for D2n+1R=90cmμ=1Find rn=?D2nD2nD2n+1=4Rnλ=4Rnλ1=4R(n+1)λ2D2n4Rnλ1n=D2n+1=4R(n+1)λ2=λ1λ1−λ2=4500×10−8(6000−4500)×10−8=3

λ1=6000A0=6000×10−8cm for D2nλ2=4500A0=4500×10−8cm for D2n+1R=90cmμ=1Find rn=?D2nD2nD2n+1=4Rnλ=4Rnλ1=4R(n+1)λ2D2n4Rnλ1n=D2n+1=4R(n+1)λ2=λ1λ1−λ2=4500×10−8(6000−4500)×10−8=3D2n=4×90×3×6000×10−8=648×10−4

λ1=6000A0=6000×10−8cm for D2nλ2=4500A0=4500×10−8cm for D2n+1R=90cmμ=1Find rn=?D2nD2nD2n+1=4Rnλ=4Rnλ1=4R(n+1)λ2D2n4Rnλ1n=D2n+1=4R(n+1)λ2=λ1λ1−λ2=4500×10−8(6000−4500)×10−8=3D2n=4×90×3×6000×10−8=648×10−4Dnrn=25.45×10−2 cm=2545 cm=0.504 cm

λ1=6000A0=6000×10−8cm for D2nλ2=4500A0=4500×10−8cm for D2n+1R=90cmμ=1Find rn=?D2nD2nD2n+1=4Rnλ=4Rnλ1=4R(n+1)λ2D2n4Rnλ1n=D2n+1=4R(n+1)λ2=λ1λ1−λ2=4500×10−8(6000−4500)×10−8=3D2n=4×90×3×6000×10−8=648×10−4Dnrn=25.45×10−2 cm=2545 cm=0.504 cm

∴ Radius of nthdark ring = 0.504 cm

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Answer 2

Answer-0.25m

Explanation: