The width of one of the two slits in Young's double slit experiment is d while that of the other slit is x d. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9: 4 then what is the value of x ? (Assume that the field strength varies according to the slit width.)

Solution

Physics sir · 4 steps

The slit width sets the field amplitude, so add and subtract the amplitudes before you square them.

1. Amplitudes from widths. The field strength is proportional to the slit width. If the narrow slit gives amplitude E, the wider slit gives xE.
2. Bright and dark. At a maximum the two waves add in phase, so the amplitude is (x+1)E. At a minimum they cancel as far as they can, giving (x-1)E. Intensity goes as amplitude squared.

   Iₘₐₓ/Iₘᵢₙ = ((x+1)/(x-1))² = 9/4

3. Take the square root. (x+1)/(x-1) = 3/2, so 2x + 2 = 3x - 3, which gives x = 5. Check: (6/4)² = 9/4.
4. So. The wider slit is 5 times the narrow one, so x = 5, option (d).

The idea to keep. In interference, amplitudes add, not intensities. Find each amplitude first, add and subtract them for the maxima and minima, then square to get intensity.

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