If the distance from the central fringe to a first-order fringe is measured to be 2.3 mm on the screen, determine the light's wavelength (in nm).

460 nm

Explanation:

Using the formula below;

Y=B tan θ(1).

Making θ the subject of the formula, then, we have equation (2).

θ= (tan inverse) Y/B(2).

The parameters given from the question are: Y= 2.3mm= 2.3×10^-3m, B= 2.5m.

Therefore, slotting in the above parameters into the equation (2) above, we have;

θ= (tan inverse) of {2.3×10^-3/2.5}.

θ= (tan inverse) of 0.00092.

θ= 0.00092°.

Hence, dSinθ=mλ.

Therefore, λ= dsinθ(3).

Making wavelength, λ the subject of the formula. We have;

Wavelength, λ=0.5 × 10^-3 ×sin(9.2×10^-4).

λ=4.6×10^-7 m.

λ= 460nm.

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2