Final answer:
In Young's double-slit experiment, the 10th order maximum observed for a wavelength of 7000 AÅ corresponds to the 14th order maximum for a wavelength of 5000 AÅ. By setting up a ratio of the two conditions for maxima, we find that the order of maximum at the same point of observation changes to 14th, which is option B.
Step-by-step explanation:
In Young's double-slit experiment, the condition for constructive interference (maxima) at a point on the screen is given by:
d sin(θ) = mλ,
where d is the separation between the slits, θ is the angle of diffraction, m is the order of maximum, and λ is the wavelength of the light used.
For the 10th order maxima with the first wavelength (7000 Å), we have:
d sin(θ) = 10 × 7000 Å.
Using the same angle θ for the second wavelength (5000 Å), we can find the new order of maximum m
d sin(θ) = m × 5000 Å.
To find the new order m, we can set up the ratio of the two equations:
m × 5000 Å / 10 × 7000 Å = 1,
Solving for m gives us
m = 14.
Hence, at the same point of observation, the 10th order maxima for the wavelength of 7000 Å corresponds to the 14th order maxima for the wavelength of 5000 Å, which matches option B.