# A beam of light consisting of two wavelength $6500 A^{\circ}$ and $5200 A^{\circ}$ is used to obtain interference fringes in a young's double slit exp. Then what is the distance from central maxima where the bright fringes due to both wavelength coincides? The distance between slits is 2 mm and distance between plane of slits and screen is 120 cm

$(a)\;0.2\;cm \\ (b)\;0.156\;cm \\ (c)\;0.3\;cm \\ (d)\;0.9 \;cm$

If n is least number of fringes of $\lambda_1(=6500 A^{\circ})$ which are coincident with $(n+1)$ of smaller wavelength $\lambda_2=(5200 A^{\circ})$
$y' =\eta \beta =(\eta+1) \beta'$
ie $\large\frac{\eta+1}{\eta}=\frac{\beta}{\beta'}=\frac{\lambda_1}{\lambda_2}$
or $\eta =\large\frac{\lambda_2}{\lambda_1- \lambda_2} =\frac{5200}{6500-5200}$
$\quad= \large\frac{5200}{1300}$
$\quad=4$
also $\beta = \frac{\lambda D}{d}$
substituting values for wavelength,D and d we get $\beta=0.039 cm$
So, $y' =4 \beta = 4 \times 0 .039 =0.156\;cm$
Hence b is the correct answer.

edited Jul 22, 2014