To find the radius of curvature \( R \) of the convex lens using the information provided about Newton's rings, we can use the formula for the radii of the bright rings in the Newton's rings interference pattern.
The radius of the \( m \)-th bright ring in the Newton's rings setup is given by the formula:
\[ r_m^2 = m \lambda R \]
where
- \( r_m \) is the radius of the \( m \)-th bright ring,
- \( m \) is the ring number,
- \( \lambda \) is the wavelength of the light, and
- \( R \) is the radius of curvature of the lens.
In your case:
- The wavelength \( \lambda = 668 \) nm = \( 668 \times 10^{-9} \) m,
- The radius of the fiftieth bright ring \( r_{50} = 8.6 \) mm = \( 8.6 \times 10^{-3} \) m,
- The ring number \( m = 50 \).
Substituting into the formula:
\[ (8.6 \times 10^{-3})^2 = 50 \cdot (668 \times 10^{-9}) \cdot R \]
Now we calculate \( (8.6 \times 10^{-3})^2 \):
\[ (8.6 \times 10^{-3})^2 = 7.396 \times 10^{-5} \text{ m}^2 \]
Thus, we can write:
\[ 7.396 \times 10^{-5} = 50 \cdot (668 \times 10^{-9}) \cdot R \]
Now calculate \( 50 \cdot (668 \times 10^{-9}) \):
\[ 50 \cdot (668 \times 10^{-9}) = 33.4 \times 10^{-6} = 3.34 \times 10^{-5} \]
Now we can substitute this back into the equation for \( R \):
\[ 7.396 \times 10^{-5} = 3.34 \times 10^{-5} \cdot R \]
To solve for \( R \):
\[ R = \frac{7.396 \times 10^{-5}}{3.34 \times 10^{-5}} \approx 2.21 \text{ m} \]
Thus, the effective radius of curvature \( R \) of the lens is approximately:
\[ \boxed{2.21 \text{ meters}} \]
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