Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is 2.7 m. Consider a point on the screen 1.3 cm from the central maximum. Calculate (a) θ for that point, (b) α, and (c) the ratio of the intensity at that point to the intensity at the central maximum.

0.276

0.9

0.756

Explanation:

Given that

Wavelength of the light, λ = 588 nm

Distance from the slit to the screen, L = 2.7 m

Width of the slit, a = 0.0351 mm

a point on the screen, y = 1.3 cm = 0.013 m

Sinθ = y/L

Sinθ = 0.013/2.7

sinθ = 0.0081

θ = sin^-1 0.00481

θ = 0.276°

α = (π.a.sinθ)/λ

α = (3.142 * 3.51*10^-5 * sin 0.276) / 588*10^-9

α = 5.3*10^-7 / 588*10^-9

α = 0.9 rad

I/i(m) = ((sinα)/α)²

I/I(m) = ((sin 0.9) / 0.9)²

I/I(m) = (0.783/0.9)²

I/I(m) = 0.87²

I/I(m) = 0.756

Note, our calculator has to be set in Rad instead of degree for part C, to get the answer

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation:

radiation warms things up temp rises convection currents air circ windy weather