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(D) For a wavelength $\lambda$,the condition for the $n$-th maximum is path difference $\Delta x = n\lambda$,and the condition for the $m$-th minimum

Vedclass · Accepted Answer

$1500$

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(D) For a wavelength $\lambda$,the condition for the $n$-th maximum is path difference $\Delta x = n\lambda$,and the condition for the $m$-th minimum is path difference $\Delta x = (m + 1/2)\lambda$.Let the path difference be $\Delta x$. The condition for the interference pattern to be indistinct is that a maximum of $\lambda_1$ coincides with a minimum of $\lambda_2$,or vice versa.Case $1$: Maximum of $\lambda_1$ coincides with minimum of $\lambda_2$:$\Delta x = n\lambda_1 = (m + 1/2)\lambda_2$$n(500) = (m + 1/2)(600)$$5n = 6m + 3$For the smallest integer values,if $m = 1$,$5n = 9$ (no integer $n$). If $m = 4$,$5n = 27$ (no). If $m = 1$,$5n = 9$. Let's test values: $5n - 6m = 3$. For $m=1, 5n=9$; $m=2, 5n=15 \Rightarrow n=3$.Thus,$\Delta x = 3 	imes 500 = 1500\,nm$.Case $2$: Maximum of $\lambda_2$ coincides with minimum of $\lambda_1$:$\Delta x = n\lambda_2 = (m + 1/2)\lambda_1$$n(600) = (m + 1/2)(500)$$6n = 5m + 2.5 \Rightarrow 12n = 10m + 5$.Since $12n$ is even and $10m+5$ is odd,this case is impossible for integer $n, m$.Therefore,the first point of indistinctness occurs at $\Delta x = 1500\,nm$.

Similar Questions

In the Young's double slit experiment,the distance between the slits varies in time as $d(t) = d_{0} + a_{0} \sin \omega t$; where $d_{0}$,$\omega$,and $a_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:

Which of the following statements is incorrect regarding interference fringes?

In a $YDSE$,if the slits are of unequal width:

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