In a Young's double slit experiment, if the incident light consists of two wavelengths $${{\lambda }_{1}}$$and $${{\lambda }_{2}}$$, the slit separation is d, and the distance between the slit and the screen is D, the maxima due to the two wavelengths will coincide at a distance from the central maxima, given by:

Question

TuteeHUB · Accepted Answer

HCF of $$\frac{{{\lambda }_{1}}D}{d}$$ and $$\frac{{{\lambda }_{2}}D}{d}$$

TuteeHUB · Answer

$$\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{2\,Dd}$$

TuteeHUB · Answer

$$({{\lambda }_{1}}-{{\lambda }_{2}}).\frac{2d}{D}$$

TuteeHUB · Answer

LCM of $${{\lambda }_{1}}.\frac{D}{d}$$ and $${{\lambda }_{2}}.\frac{D}{d}$$

1.	In a Young's double slit experiment, if the incident light consists of two wavelengths \[{{\lambda }_{1}}\]and \[{{\lambda }_{2}}\], the slit separation is d, and the distance between the slit and the screen is D, the maxima due to the two wavelengths will coincide at a distance from the central maxima, given by:
A.	\[\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{2\,Dd}\]
B.	\[({{\lambda }_{1}}-{{\lambda }_{2}}).\frac{2d}{D}\]
C.	LCM of \[{{\lambda }_{1}}.\frac{D}{d}\] and \[{{\lambda }_{2}}.\frac{D}{d}\]
D.	HCF of \[\frac{{{\lambda }_{1}}D}{d}\] and \[\frac{{{\lambda }_{2}}D}{d}\]
Answer» D. HCF of \[\frac{{{\lambda }_{1}}D}{d}\] and \[\frac{{{\lambda }_{2}}D}{d}\]

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