The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x : 4 where x is ____________.

Solution

Physics sir · 3 steps

Width sets the amplitude here, so take amplitudes a and 3a, then add and subtract them.

1. Amplitudes. The amplitude is proportional to slit width. So if the narrow slit gives amplitude a, the wide slit gives 3a.
2. Bright and dark. At a bright fringe the amplitudes add, giving 4a. At a dark fringe they subtract, giving 2a. Intensity is the square of the amplitude.

   Iₘᵢₙ/Iₘₐₓ = (3a - a)²/(3a + a)² = 4a²/16a² = 1/4

3. So. Iₘᵢₙ : Iₘₐₓ = 1 : 4, so x = 1.

The common slip: 2. Forgetting to square the amplitudes gives (3a - a)/(3a + a) = 2/4, so x = 2. Intensity is amplitude squared.

The idea to keep. In interference, amplitudes add and subtract, never intensities. Find A₁ ± A₂ first, then square both to get Iₘₐₓ and Iₘᵢₙ.

Answer from the NTA final answer key. Solution by PrepWiser.