Final answer:
In Young's double-slit experiment, the product of the fringe width and number of fringes for a longer wavelength will be less than that for a shorter wavelength. Hence, w1 * n1 < w2 * n2. The correct answer is: c) w₁ * n₁ < w₂ * n₂.
Step-by-step explanation:
In Young's double-slit experiment, the fringe width (w) is directly proportional to the wavelength (λ) and inversely proportional to the slit separation. The number of fringes (n) within a distance (D) on one side of the central maximum is given by the equation n = D / w. To compare the product of the fringe width and the number of fringes for two different wavelengths, λ₁ and λ₂, with the recorded fringe widths w₁ and w₂ and the number of fringes n₁ and n₂, we use the relationship w = λD / d, where d is the slit separation and D is the distance from the slits to the screen.
- If λ₁ > λ₂, then w₁ > w₂, assuming the distance (D) and the slit separation (d) remain constant for both wavelengths.
- However, since the number of fringes is found by dividing the distance (D) by the fringe width, n₁ will be less than n₂ because w₁ > w₂ and n = D / w.
- Therefore, the product w₁ * n₁ will be less than w₂ * n₂, because the increase in w₁ relative to w₂ is exactly offset by the decrease in n₁ relative to n₂.
Hence, the correct answer to the student's question is: c) w₁ * n₁ < w₂ * n₂.