Newton's rings are observed. normally in reflected light of wavelength 5000 mathring A . The diameter of the 10th dark ring is 0.005 m. The radius of curvature of the lens is

Question

Newton's rings are observed. normally in reflected light of wavelength 5000 mathring A . The diameter of the 10th dark ring is 0.005 m. The radius of curvature of the lens is (A) 1 m (B) 1.25 m (C) 2 m (D) 0.5 m

Tardigrade Admin · Accepted Answer

Diameter of nth dark ring, Dn = 2 √n λ R ∴ R = (D2n/4n λ) Here, n = 10, D10 = 0.005 : m λ = 5000 mathring A = 5000 × 10-10 m = 5× 10-7 m ∴ R= ((0.005 m)2/4× 10× 5× 10-7m) = 1.25 m

Q. Newton's rings are observed. normally in reflected light of wavelength $5000 \overset{˚}{A}$ . The diameter of the 10 $^{t h}$ dark ring is 0.005 m. The radius of curvature of the lens is

1 m

1.25 m

2 m

0.5 m

Diameter of $n^{t h}$ dark ring,
$D_{n} = 2 nλ R$
$∴ R = \frac{D _{n}^{2}}{4 nλ}$
Here, $n = 10, D_{10} = 0.005 m$
$λ = 5000 \overset{˚}{A} = 5000 \times 1 0^{- 10} m = 5 \times 1 0^{- 7} m$
$∴ R = \frac{( 0.005 m ) ^{2}}{4 \times 10 \times 5 \times 1 0 ^{- 7} m} = 1.25 m$

Q. Newton's rings are observed. normally in reflected light of wavelength 5000A˚ . The diameter of the 10th dark ring is 0.005 m. The radius of curvature of the lens is

1 m

1.25 m

2 m

0.5 m

Diameter of nth dark ring, Dn​=2nλR​ ∴R=4nλDn2​​ Here, n=10,D10​=0.005m λ=5000A˚=5000×10−10m=5×10−7m ∴R=4×10×5×10−7m(0.005m)2​=1.25m

Q. Newton's rings are observed. normally in reflected light of wavelength $5000 \overset{˚}{A}$ . The diameter of the 10 $^{t h}$ dark ring is 0.005 m. The radius of curvature of the lens is

Diameter of $n^{t h}$ dark ring,
$D_{n} = 2 nλ R$
$∴ R = \frac{D _{n}^{2}}{4 nλ}$
Here, $n = 10, D_{10} = 0.005 m$
$λ = 5000 \overset{˚}{A} = 5000 \times 1 0^{- 10} m = 5 \times 1 0^{- 7} m$
$∴ R = \frac{( 0.005 m ) ^{2}}{4 \times 10 \times 5 \times 1 0 ^{- 7} m} = 1.25 m$