Question

A Newton's ring apparatus consists of a convex lens with a large radius of curvature \(R\) placed on a flat glass disc. (a) Show that the horizontal distance \(x\) from the center, the thickness \(d\) of the air gap, and the radius of curvature \(R\) are related by \(x^{2}=2 R d\). (b) Show that the radius of the \(n\) th constructive interference ring is given by \(x_{n}=\left[\left(n+\frac{1}{2}\right) \lambda R\right]^{1 / 2}\). (c) How many bright rings may be seen if the apparatus is illuminated by red light of wavelength \(700 . \mathrm{nm}\) with \(R=10.0 \mathrm{~m}\) and the plane glass disc diameter \(5.00 \mathrm{~cm} ?\)

Short answer

Answer: The relationship between the horizontal distance x, the thickness of the air gap d, and the radius of curvature R is given by the approximation x² ≈ 2Rd. The radius of the nth constructive interference ring can be determined using the expression x_n = [(n + 1/2)λR]^(1/2), where λ is the wavelength of the light. To find the number of bright rings visible, we can use this expression and compare it to the radius of the plane glass disc, solving for n, which gives us the maximum number of visible rings. In this case, n = 12 rings can be seen when the apparatus is illuminated by red light.

Step-by-step solution

(a) Relationship between x, d, and R

We can use geometry to find the relationship between x, d, and R. Consider a cross-sectional view of the convex lens and the flat glass disc on which the lens is placed. Let us draw a right triangle with vertices at the point where the lens contacts the disc (A), the center of the lens curvature (C), and any point along the horizontal distance x (B), from the center of the lens. In triangle ABC, we have: 1. AC = R (radius of curvature) 2. BC = x (horizontal distance from the center) 3. AB = R - (R - d) = d (the thickness of the air gap) According to Pythagorean theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In triangle ABC, we have AC² = AB² + BC². Plugging in the variables, we get: \[R^2 = d^2 + x^2\] Now, we can solve for d: \[d = \sqrt{R^2 - x^2}\] Since d is typically small compared to R, we can use the binomial approximation, which states that for small x and y, (1 + x)^y ≈ 1 + xy. In our case, we have: \[d \approx \sqrt{R^2 - x^2} = R \sqrt{1 - \frac{x^2}{R^2}} \approx R\left(1 - \frac{x^2}{2R^2}\right)\] Using this approximation, we can solve for x²: \[x^2 \approx 2Rd\]

(b) Radius of the nth constructive interference ring

In order to find the radius of the nth constructive interference ring, we need to apply the condition for constructive interference. The constructive interference occurs when the path difference between the rays reflecting off the lens surface and the rays reflecting off the bottom surface of the air gap is an integer multiple of the wavelength λ. Let's denote the path difference as ΔL, then for constructive interference: \[\Delta L = n\lambda\] where n is an integer. The path difference ΔL is twice the air gap thickness d, since the rays reflecting off the lens surface and the rays reflecting off the bottom surface of the air gap travel twice the distance d. Hence, \[2d = n\lambda\] Using the relationship we derived in part (a), we can substitute the expression for d into this equation: \[2(2Rd) \approx n\lambda\] Now, we can solve for x_n: \[x_n = \sqrt{2R(n\lambda)}\] In the problem statement, we are given that we need to show the radius of the nth constructive interference ring is given by: \[x_n = \left[\left(n + \frac{1}{2}\right)\lambda R\right]^{1/2}\] This can simply be achieved by adding \(0.5\lambda\) to both sides of the equation we derived above: \[x_n = \sqrt{2R(n\lambda + 0.5\lambda)} = \left[\left(n + \frac{1}{2}\right)\lambda R\right]^{1/2}\]

(c) Number of bright rings

To find the number of bright rings, we'll use the expression we derived for the radius of the nth constructive interference ring and the given values for the wavelength (700 nm), the radius of curvature (10 m), and the plane glass disc diameter (5 cm). The radius of the plane glass disc is half of its diameter, which is 2.5 cm or 0.025 m. The nth bright ring must be within this radius to be visible. Therefore, \[x_n = \left[\left(n + \frac{1}{2}\right)(700 \times 10^{-9})(10)\right]^{1/2} \le 0.025\] Squaring both sides, we get: \[\left(n + \frac{1}{2}\right)(700 \times 10^{-9})(10) \le 0.025^2\] Now, we solve for n: \[n \le \frac{0.025^2}{(700 \times 10^{-9})(10)} - \frac{1}{2}\] Evaluating the expression, we get: \[n \le 12.99\] Since n must be an integer, the maximum value for n is 12. Hence, there will be 12 bright rings visible with the given parameters.