In Young's experiment,the separation between the $5^{\text{th}}$ maxima and the $3^{\text{rd}}$ minima is how many times the fringe width?

  • A
    $5$
  • B
    $3$
  • C
    $2.5$
  • D
    $2$

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Similar Questions

Two slits separated by a distance of $1\, mm$ are illuminated with red light of wavelength $6.5 \times 10^{-7}\, m$. The interference fringes are observed on a screen placed $1\, m$ from the slits. Find the distance between the third dark fringe and the fifth bright fringe on the same side of the central maxima.

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In a Young's double slit experiment, the slits are separated by $0.3 \, mm$ and the screen is $1.5 \, m$ away from the plane of slits. The distance between the fourth bright fringes on both sides of the central bright fringe is $2.4 \, cm$. The frequency of light used is $.......... \times 10^{14} \, Hz$.

Which of the following statements is incorrect regarding interference fringes?

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-$II$
$(A) x = \frac{D \lambda}{d}$ $(P) I_0$
$(B) x = \frac{D \lambda}{4d}$ $(Q) 2 I_0$
$(C) x = \frac{D \lambda}{3d}$ $(R) 3 I_0$
$(D) x = \frac{D \lambda}{6d}$ $(S) 4 I_0$

In a $YDSE$ experiment,if a slab whose refractive index can be varied is placed in front of one of the slits,then the variation of resultant intensity at the mid-point of the screen with $\mu$ will be best represented by $(\mu \geq 1)$. [Assume slits of equal width and there is no absorption by the slab]

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