The Young's double slit experiment is performed with monochromatic blue, green, yellow and red lights, respectively. Arrange the distance(x) of 4th maxima from the central maximum for all the colors in decreasing order
(A) \( x_{\mathrm{red}} \)
(B) \( x_{\mathrm{blue}} \)
(C) \( x_{\mathrm{green}} \)
(D) \( x_{\mathrm{yellow}} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (D), (C), (B)
In Young's double slit experiment the \( n \)th bright fringe lies at \( x_n = n\lambda D/d \). For the 4th maximum and the same slits and screen, \( x \) is proportional to the wavelength.
\( \displaystyle x_4 = \frac{4\lambda D}{d} \;\Rightarrow\; x \propto \lambda \)
In the visible spectrum the wavelength decreases from red through yellow and green to blue.
\( \displaystyle \lambda_{\mathrm{red}} \gt \lambda_{\mathrm{yellow}} \gt \lambda_{\mathrm{green}} \gt \lambda_{\mathrm{blue}} \)
So the distance in decreasing order is \( x_{\mathrm{red}} \), \( x_{\mathrm{yellow}} \), \( x_{\mathrm{green}} \), \( x_{\mathrm{blue}} \), that is (A), (D), (C), (B).
NCERT (2026-27):
Interference of Light Waves and Young’s Experiment – “… We will have constructive interference resulting in a bright region when
\( \displaystyle \frac{xd}{D} = n\lambda \)
That is,
\( \displaystyle x = x_n = \frac{n\lambda D}{d}; \quad n = 0, \pm 1, \pm 2, \ldots \qquad (10.13) \)”