Two identical narrow slits $S_1$ and $S_2$ are illuminated by light of wavelength $\lambda$ from a point source $P$. If,as shown in the diagram,the light is then allowed to fall on a screen,and if $n$ is a positive integer,the condition for destructive interference at $Q$ is that
- A$(l_1 - l_2) = (2n + 1)\lambda /2$
- B$(l_3 - l_4) = (2n + 1)\lambda /2$
- C$(l_1 + l_2) - (l_2 + l_4) = n\lambda$
- D$(l_1 + l_3) - (l_2 + l_4) = (2n + 1)\lambda /2$
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Similar Questions
In a Young's double-slit experiment with wavelength $\lambda$,the fringe width is $\beta$. When two glass plates of thicknesses $t_1$ and $t_2$ $(t_1 > t_2)$ and refractive index $\mu$ are placed in the paths of the two light beams respectively,by what distance will the fringe pattern shift?
Medium
View SolutionIn $\text{YDSE}$,$S_1$ 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-$II$ $(A) x = \frac{D \lambda}{d}$ $(P) I_0$ $(B) x = \frac{D \lambda}{4d}$ $(Q) 2 I_0$ $(C) x = \frac{D \lambda}{3d}$ $(R) 3 I_0$ $(D) x = \frac{D \lambda}{6d}$ $(S) 4 I_0$
| Column-$I$ | Column-$II$ |
|---|---|
| $(A) x = \frac{D \lambda}{d}$ | $(P) I_0$ |
| $(B) x = \frac{D \lambda}{4d}$ | $(Q) 2 I_0$ |
| $(C) x = \frac{D \lambda}{3d}$ | $(R) 3 I_0$ |
| $(D) x = \frac{D \lambda}{6d}$ | $(S) 4 I_0$ |
Medium
View SolutionTwo slits separated by a distance of $1 \,mm$ are illuminated with light of wavelength $6.5 \times 10^{-7} \,m$. The interference fringes are observed on a screen placed at $1 \,m$ from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (in $\,mm$)
The graph shows the variation of fringe width $(X)$ versus the distance of the screen from the plane of the slits $(D)$ in Young's double-slit experiment (keeping other parameters constant,where $d$ is the distance between the slits). The wavelength of light used can be calculated as:
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......$mm$.
DifficultIIT 2004
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