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Q10.6    A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

            (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

Answers (1)

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Given,

The wavelength of one light beam  :

 \lambda_1=650nm=650*10^{-9}m  

The wavelength of another Light beam

\lambda_2=520nm=520*10^{-9}m

Let, the distance between the two-slit be d and distance between slit and screen is D

NOW,

As we know, the distance x of nth bright fringe from central maxima is given by 

x=n\lambda \frac{D}{d}

so for 3rd fringe,

n=3

x=n\lambda_1 \frac{D}{d}=3*650*10^{-9}*\frac{D}{d}=1950\frac{D}{d}nm

Hence distance of 3rd fringe from central maxima is    1950\frac{D}{d}nm. Here value D and d are not given in the question.

Posted by

Pankaj Sanodiya

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