Problems on Newton’s Rings Experiment – Solved Numericals

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1. Newton’s rings are observed in the reflected light of wavelength 5900 ˚A. The diameter of 10th dark ring is 0.5 cm. Find the radius of curvature of the lens used.

Solution:

• Wavelength of light, \( \lambda = 5900 \, \text{Å} = 5900 \times 10^{-10} \, \text{m} = 5.9 \times 10^{-7} \, \text{m} \)
• Diameter of the 10th dark ring, \( D_{10} = 0.5 \, \text{cm} = 0.005 \, \text{m} \)
• Number of the dark ring, \( n = 10 \)
• We need to find the radius of curvature of the lens, \( R \).

The diameter of the \( n \)-th dark ring is given by:

\[ D_n^2 = 4 n \lambda R \] \[ R = \frac{D_n^2}{4 n \lambda} \] \[ R = \frac{(0.005)^2}{4 \times 10 \times 5.9 \times 10^{-7}} \] \[ R = 1.059 \text{ m} \]

2. Calculate the thickness of air film at the 10th dark ring in a Newton’s rings system, viewed normally by a reflected light of wavelength 500 nm. The diameter of the 10^th dark ring is 2 mm.

Solution:

• Wavelength of light, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)
• Diameter of the 10th dark ring, \( D_{10} = 2 \, \text{mm} = 2 \times 10^{-3}= 0.002 \, \text{m} \)
• Number of the dark ring, \( n = 10 \)
• We need to find the thickness of the air film, \( t \), at the position of the \(10^{th}\) dark ring.

\[ R = \frac{D_n^2}{4 n \lambda} \] \[ R = \frac{(0.002)^2}{4 \times 10 \times 5 \times 10^{-7} } \] \[ R = \frac{4\times 10^{-6} }{2 \times 10^{-5} } \] \[ R = 0.2 \text{ m} \]

Thicknicess of the film

Radius of the ring, \[r_{10} = \frac{D_{10}}{2} = \frac{0.002}{2} = 0.001 \]

\[ t = \frac{r_{10}^2}{2R} \] \[ t = \frac{(0.001)^2}{2\times 0.2} \] \[ t = \frac{1\times 10^{-6}}{2\times 0.2} = 2.5 \times 10^{-6} \] \[ \] \[ t = 2.5 \mu m \]

3. A plano-convex lens of radius of curvature R=1 m is placed on a glass plate. The wavelength of light is used 500 nm. Calculate the radius of the 5th dark ring observed in the pattern of Newton's rings.

Solution:

• Radius of curvature of the plano-convex lens, \( R = 1 \, \text{m} \)
• Wavelength of light, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)
• Number of the dark ring, \( n = 5 \)
• We need to find the radius of the 5th dark ring, \( r_5 \).

\[ D_n^2 = 4 n \lambda R \] \[ D_5^2 = 4 \times 5 \times 5 \times 10^{-7} \times 1 \] \[ D_5^2 = 10^{-5} \times 1 \] \[ D_5 = 3.16 \times 10^{-3} \times 1 \] \[ r_5= 1.58 \times 10^{-3} \] \[ r_5= 1.58 \text{ mm} \]

4. In Newton's ring experiment the diameter of the 15th ring was found to be 0.59 cm and that of the 5th ring was 0.336 cm. If the radius of the plano-convex lens is 100 cm, compute the wavelength of light used.

Solution:

• Diameter of the 15th dark ring, \( D_{15} = 0.59 \, \text{cm} = 0.0059 \, \text{m} \)
• Diameter of the 5th dark ring, \( D_5 = 0.336 \, \text{cm} = 0.00336 \, \text{m} \)
• Radius of curvature of the plano-convex lens, \( R = 100 \, \text{cm} = 1 \, \text{m} \)
• We need to find the wavelength of light, \( \lambda \).

\[ \lambda = \frac{D_{m}^2-D_{n}^2}{4\times (m-n) \times R} \] \[ \lambda = \frac{D_{15}^2-D_{5}^2}{4\times 10 \times 1} \] \[ \lambda = \frac{(0.0059)^2-(0.00336)^2}{4\times 10 \times 1} \] \[ \lambda = \frac{3.481\times 10^{-5}-1.12896\times 10^{-5}}{40} = 0.0588 \times 10^{-5} \] \[ \lambda = 588 \text{ nm} \]