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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Wave Optics

In YDSE if the source consists of two wave lengths $\lambda_1=4000 \;a^{\circ}$ and $\lambda _2 =4002\;A^{\circ}$ . Find the distance from centre where fringes disappear if $d= 1\;cm; D=1\;m$

$(a)\;20\;mm \\ (b)\;40\;mm \\ (c)\;60\;mm \\ (d)\;80\;mm $

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1 Answer

The fringes disappear when the maxima of $ \lambda_1$ fall over minima of $\lambda_2$.
$\large\frac{p}{\lambda_1}-\frac{p}{\lambda_2} =\frac{1}{2}$
Where p is the optical path difference at that point.
or $p=\large\frac{\lambda_1 \lambda_2}{2( \lambda_2 - \lambda_1)}$
Here $\lambda_1=4000\;A^{\circ}, \lambda_2= 4002 A^{\circ}$
$\therefore p=0.04\;cm$
in YDSE
$p= \large\frac{d y}{D}$
$\therefore y= \large\frac{D}{d}$$p$
$\qquad= \large\frac{(1000)}{10}$$(0.4) =40\;mm$
Hence b is the correct answer.
answered Feb 20, 2014 by meena.p
 

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