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Home » In Young’s double slit experiment, in an interference pattern second minimum is observed exactly in front of one slit. The distance between the two coherent sources is ′d′ and the distance between source and screen is ′D′. The wavelength of light source used is :
In Young’s double slit experiment, in an interference pattern second minimum is observed exactly in front of one slit. The distance between the two coherent sources is ′d′ and the distance between source and screen is ′D′. The wavelength of light source used is :
Physics | Physics
In Young’s double slit experiment, in an interference pattern second minimum is observed exactly in front of one slit. The distance between the two coherent sources is ′d′ and the distance between source and screen is ′D′. The wavelength of light source used is :

 

A. d2/D

B. d2/2D

C. d2/3D

D. d2/4D

Solution: The correct answer is C. d2/3D

Because the second minima occur right in front of one of the slits, y = d/2

For\text{ }second\text{ }minima~(n=1),~

y=\left( n+\frac{1}{2} \right)\frac{\lambda D}{d}

where~\lambda ~is\text{ }the\text{ }wavelength\text{ }of\text{ }light\text{ }used.

\frac{d}{2}=\left( 1+\frac{1}{2} \right)\frac{\lambda D}{d}

\frac{d}{2}=\frac{3\lambda D}{2d}

\lambda =\frac{{{d}^{2}}}{3D}

 

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