The diameter of the 9th dark ring in the newton’s ring experiment is 0.28 cm. What is the diameter of
16th dark ring? Given the wavelength of light used is 6000 A0
. Calculate the radius of curvature of the
lens used.
Dn=0.28cm=0.0028mD_n = 0.28 cm = 0.0028 mDn=0.28cm=0.0028m
λ=6000A0=6000×10−10m\lambda = 6000 A^0 = 6000 \times 10^{-10}mλ=6000A0=6000×10−10m
Radius of curvature
R=Dn24nλR = \frac {D^2_n}{4n \lambda}R=4nλDn2
R=0.002824×9×6000×10−10=0.3629m=36.29cmR = \frac {0.0028^2}{4 \times 9 \times 6000 \times 10^{-10}} = 0.3629 m = 36.29 cmR=4×9×6000×10−100.00282=0.3629m=36.29cm
diameter of 16th dark ring
D16=R×4nλD_{16} = \sqrt{R \times 4 n \lambda}D16=R×4nλ
D16=0.3629×4×16×6000×10−10=0.003733m=0.3733cmD_{16} = \sqrt{0.3629 \times 4\times16 \times 6000 \times 10^{-10}} = 0.003733 m =0.3733 cmD16=0.3629×4×16×6000×10−10=0.003733m=0.3733cm
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