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(C) The position of the $n^{th}$ bright fringe is given by $y = n \frac{D \lambda}{d}$.For the first bright fringe $(n = 1)$,the position is $y = \fra

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(C) The position of the $n^{th}$ bright fringe is given by $y = n \frac{D \lambda}{d}$.For the first bright fringe $(n = 1)$,the position is $y = \frac{D \lambda}{d}$.Given $D = 1.5 \, m$,$d = 0.3 \, mm = 0.3 	imes 10^{-3} \, m$.For red light,$y_r = 3.5 \, mm = 3.5 	imes 10^{-3} \, m$. Thus,$\lambda_r = \frac{y_r d}{D} = \frac{3.5 	imes 10^{-3} 	imes 0.3 	imes 10^{-3}}{1.5} = 0.7 	imes 10^{-6} \, m = 700 \, nm$.For violet light,$y_v = 2.0 \, mm = 2.0 	imes 10^{-3} \, m$. Thus,$\lambda_v = \frac{y_v d}{D} = \frac{2.0 	imes 10^{-3} 	imes 0.3 	imes 10^{-3}}{1.5} = 0.4 	imes 10^{-6} \, m = 400 \, nm$.The difference in wavelengths is $\Delta \lambda = \lambda_r - \lambda_v = 700 \, nm - 400 \, nm = 300 \, nm$.

Similar Questions

In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.

In Young's double-slit experiment,when two light waves form the third minimum,the path difference between them is:

Statement-$1$: If white light is used in $YDSE$,then the central bright fringe will be white.
Statement-$2$: In the case of white light used in $YDSE$,all the wavelengths produce their zero-order maxima at the same position.

In a double slit experiment,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. The fringe width:

When a thin metal plate is placed in the path of one of the interfering beams of light in a Young's double-slit experiment,what happens to the interference pattern?

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