If white light is used in Young's double-slit experiment, then a very large number of coloured fringes can be seen. Which of the following statements is correct regarding the fringe pattern?

  • A
    With first-order violet fringes being closer to the central white fringe.
  • B
    First-order red fringes being closer to the central white fringe.
  • C
    With a central white fringe.
  • D
    With a central black fringe.

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Similar Questions

In a Young's double-slit experiment,the light beam consists of two wavelengths $6500 \, \mathring A$ and $5200 \, \mathring A$. The distance between the slits is $2 \, mm$ and the distance between the plane of the slits and the screen is $120 \, cm$. Find the distance of the third bright fringe from the central maximum for the wavelength $6500 \, \mathring A$ in $mm$.

The angular width of the central maximum in a single slit diffraction pattern is $60^o$. The width of the slit is $1 \mu m$. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it,Young's fringes can be observed on a screen placed at a distance $50 \ cm$ from the slits. If the observed fringe width is $1 \ cm$,what is the slit separation distance in $\mu m$ (i.e.,distance between the centres of each slit)?

In $\text{YDSE}$,$S_1$ and $S_2$ have the same intensity $I_0$. Column-$I$ shows the distance $x$ of a point $P$ from the central point $O$ on the screen,and Column-$II$ shows the intensity at $P$. Match Column-$I$ with Column-$II$. (Wavelength is $\lambda$)
Column-$I$ Column-$II$
$(A) x = \frac{D \lambda}{d}$ $(P) I_0$
$(B) x = \frac{D \lambda}{4d}$ $(Q) 2 I_0$
$(C) x = \frac{D \lambda}{3d}$ $(R) 3 I_0$
$(D) x = \frac{D \lambda}{6d}$ $(S) 4 I_0$

Assertion: In Young's double slit experiment,the two slits are at a distance $d$ apart. An interference pattern is observed on a screen at a distance $D$ from the slits. At a point on the screen directly opposite to one of the slits,a dark fringe is observed. Then,the wavelength of the wave is proportional to the square of the distance between the two slits.
Reason: For a dark fringe,the intensity is zero.

In a $YDSE$,bi-chromatic light of wavelengths $400 \, nm$ and $560 \, nm$ are used. The distance between the slits is $0.1 \, mm$ and the distance between the plane of the slits and the screen is $1 \, m$. The minimum distance between two successive regions of complete darkness is......$mm$.

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