Final answer:
The angle of deviation for the 1st order maxima of monochromatic light with a wavelength of 440 nm passing through slits 6.2 μm apart is approximately 4.07°.
Step-by-step explanation:
The angle of deviation for the 1st order maxima in Young's Double Slit Experiment can be determined using the formula for constructive interference:
d sin(θ) = mλ,
where d is the distance between the slits, θ is the angle of deviation, m is the order of the maxima, and λ is the wavelength of the light. Since we are looking at the 1st order maxima (m = 1), and the given values are λ = 440 nm (which is 440 x 10-9 meters) and d = 6.2 μm (which is 6.2 x 10-6 meters), we can rearrange the formula to solve for θ:
sin(θ) = λ / d
sin(θ) = (440 x 10-9) / (6.2 x 10-6)
θ = sin-1((440 x 10-9) / (6.2 x 10-6))
Calculating this we get:
θ ≈ 4.07°
Therefore, the angle of deviation for the 1st order maxima is approximately 4.07 degrees.