$(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.

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