Read the information given below to answer
In Young’s double slit experiment as shown in figure, interference of light waves were observed on the screen. Thomas Young made two pinholes \( S_1 \) and \( S_2 \) (very close to each other) on an opaque screen in front of parent source \( S \)
\( S_1 \) and \( S_2 \) behaved as coherent sources, and produced interference pattern on the screen which has alternate bright fringe and dark fringes. This experiment proved Huygens’ wave theory of Light.
In YDSE experiment position of bright fringes and dark fringes is given by
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( x_n = n\displaystyle\frac{\lambda D}{d} \) \( n = 0, \pm 1, \pm 2 \ldots \) \( x_n = \left(n + \displaystyle\frac{1}{2}\right)\displaystyle\frac{\lambda D}{d} \) \( n = 0, \pm 1, \pm 2 \ldots \)
At a point \( x \) from the centre the path difference is \( xd/D \). A bright fringe forms where the path difference is a whole number of wavelengths, and a dark fringe where it is an odd number of half wavelengths.
\( \displaystyle \frac{x_n d}{D} = n\lambda \;\Rightarrow\; x_n = n\,\frac{\lambda D}{d}, \quad n = 0, \pm 1, \pm 2, \ldots \)
\( \displaystyle \frac{x_n d}{D} = \left(n + \frac{1}{2}\right)\lambda \;\Rightarrow\; x_n = \left(n + \frac{1}{2}\right)\frac{\lambda D}{d}, \quad n = 0, \pm 1, \pm 2, \ldots \)
The bright-fringe set includes \( n = 0 \), the central bright fringe.
Why the other options are incorrect
Option (2)
\( x_n = n\displaystyle\frac{\lambda d}{D} \)
\( n = \pm 1, \pm 2 \ldots \)
\( x_n = \left(n - \displaystyle\frac{1}{2}\right)\displaystyle\frac{\lambda D}{d} \)
\( n = \pm 1, \pm 2 \ldots \)
— Incorrect. The bright-fringe position has \( d \) and \( D \) interchanged, and it leaves out \( n = 0 \), so the central bright fringe would be missing.Option (3)
\( x_n = n\lambda \)
\( n = 0, \pm 1, \pm 2 \ldots \)
\( x_n = \left(n + \displaystyle\frac{1}{2}\right)\lambda \)
\( n = 0, \pm 1, \pm 2 \ldots \)
— Incorrect. \( n\lambda \) is the path difference for a bright fringe, not its position on the screen.Option (4)
\( x_n = n\displaystyle\frac{\lambda}{d} \)
\( n = \pm 1, \pm 2 \ldots \)
\( x_n = \left(n + \displaystyle\frac{1}{2}\right)\displaystyle\frac{\lambda}{d} \)
\( n = \pm 1, \pm 2 \ldots \)
— Incorrect. \( n\lambda/d \) is the angular position (in radians) of a bright fringe; it is not multiplied by \( D \), so it is not a distance on the screen.
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,
\( \displaystyle x = x_n = \frac{n\lambda D}{d};\; n = 0, \pm 1, \pm 2, \ldots \qquad (10.13) \)
On the other hand, we will have destructive interference resulting in a dark region when \( \displaystyle \frac{xd}{D} = \left(n + \frac{1}{2}\right)\lambda \) that is
\( \displaystyle x = x_n = \left(n + \frac{1}{2}\right)\frac{\lambda D}{d};\; n = 0, \pm 1, \pm 2 \qquad (10.14) \)
Thus dark and bright bands appear on the screen, as shown in Fig. 10.13. Such bands are called fringes. Equations (10.13) and (10.14) show that dark and bright fringes are equally spaced.”