3. In Newton’s rings experiment, the radius of curvature R of the lens is 5.0 m and its diameter is 20 mm a) How many rings are produced? b) How many rings would be seen if the arrangement were immersed in water (n=1.33)? Assume that λ = 589 nm

To solve this problem, we need to use the formula for the radius of the mth dark ring in Newton's rings experiment, which is given by:

r_m = sqrt(mλR)

where:

• r_m is the radius of the mth dark ring
• m is the order of the ring
• λ is the wavelength of light used
• R is the radius of curvature of the lens

a) To find the number of rings produced, we first need to find the maximum order of the ring that fits within the diameter of the lens. The diameter of the lens is 20 mm, so the radius of the lens is 10 mm or 0.01 m. We can set r_m equal to this value and solve for m:

0.01 m = sqrt(m * 589 nm * 5.0 m) 0.01 m = sqrt(m * 589e-9 m * 5.0 m) Squaring both sides: (0.01 m)^2 = m * 589e-9 m * 5.0 m 1e-4 m^2 = m * 2945e-9 m^2 m = 1e-4 m^2 / 2945e-9 m^2 m ≈ 34

So, approximately 34 rings are produced.

b) When the arrangement is immersed in water, the wavelength of light changes. The new wavelength in water is given by λ/n, where n is the refractive index of water. So, the new wavelength is 589 nm / 1.33 ≈ 443 nm. We can substitute this new wavelength into the formula and solve for m again:

0.01 m = sqrt(m * 443 nm * 5.0 m) 0.01 m = sqrt(m * 443e-9 m * 5.0 m) Squaring both sides: (0.01 m)^2 = m * 2215e-9 m^2 m = 1e-4 m^2 / 2215e-9 m^2 m ≈ 45

So, approximately 45 rings would be seen if the arrangement were immersed in water.