Final answer:
The highest-order maximum for 400-nm light falling on double slits separated by 25.0 μm is the 62nd order, calculated using the formula for constructive interference and the condition that sin θ equals 1 at the maximum possible angle of 90 degrees.
Step-by-step explanation:
The question asks to find the highest-order maximum for 400-nm light falling on double slits separated by 25.0 μm. The highest-order maximum can be calculated using the formula for constructive interference in a double-slit experiment: mλ = d sin θ, where λ is the wavelength of light, d is the distance between the slits, m is the order of the maximum, and θ is the angle of the maximum from the central axis.
To find the highest-order maximum, we set sin θ = 1, which corresponds to the largest possible value of θ (90 degrees). Rearranging for m gives us m = d/λ. Substituting d = 25.0 μm = 25.0 x 10-6m and λ = 400 nm = 400 x 10-9m, we find that the highest-order m is given by:
m = (25.0 x 10-6m) / (400 x 10-9m) = 62.5
Since the order number must be an integer, we round down to the nearest whole number. Therefore, the highest-order maximum that can be observed is the 62nd order.