In Young's double Slit experiment 3rd maxima (Bright Band) is obtained at a distance \( d_1 \) from the central bright fringe when light of wavelength \( \lambda_1 \) is used and at a distance \( d_2 \) when light of wavelength \( \lambda_2 \) is used. What will be the ratio of \( d_1 \) and \( d_2 \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\lambda_1}{\lambda_2} \)
In Young's double slit experiment the \( n \)th bright fringe is at a distance
\( \displaystyle x_n = \frac{n\lambda D}{d} \)
from the central bright fringe, where \( D \) is the slit-screen distance and \( d \) the slit separation. For the third maximum (\( n = 3 \)) with the same apparatus:
\( \displaystyle d_1 = \frac{3\lambda_1 D}{d}, \qquad d_2 = \frac{3\lambda_2 D}{d} \)
\( \displaystyle \frac{d_1}{d_2} = \frac{\lambda_1}{\lambda_2} \)
The fringe positions are directly proportional to the wavelength.