Here, d = 2 mm, D = 1.2 m, $\lambda_1$ = 650 nm = 650 × $10^{-9}$ m, $\lambda_2$ = 520 nm = 520 × $10^{-9}$ m (a) Distance of third bright fringe from the central maximum for the wavelength 650 nm. $y_3$ = $\frac{3\lambda D}{d}$ = $\frac{3(650\times10^{-9})1.2}{2\times10^{-3}}$ = 1.17 mm (b) Let at linear distance ‘y’ from center of screen the bright fringes due to both wavelength coincides. Let $n_1$ number of bright fringe with wavelength $\lambda_1$ coincides with $n_2$ number of bright fringe with wavelength $\lambda_2$. We can write, y = $n_1\beta_1$ = $n_2\beta_2$ $n_1\frac{\lambda_1 D}{d}$ = $n_2\frac{D\lambda_2}{d}$ or $n_1\lambda_1$ = $n_2\lambda_2$ ... (i) Also at first position of coincide, the nth bright fringe of one will coincide with (n + 1)th bright fringe of other. If $\lambda_2$ < $\lambda_1$ So, then $n_2$ > $n_1$ then $n_2$ = $n_1+1$ ... (ii) Using equation (ii) in equation (i) $n_1\lambda_1$ = $(n_1 + 1)\lambda_2$ ⇒ $n_1$ (650) × $10^{-9}$ = $(n_1 + 1) 520 \times 10^{-9}$ 65 $n_1$ = 52 $n_1$ + 52 or 12 $n_1$ = 52 or $n_1$ = 4 Thus, y = $n_1\beta_1$ = 4 $\left[\frac{6.5\times10^{-7}(1.2)}{2\times10^{-3}}\right]$ = 1.56 × $10^{-3}$ m = 1.56 mm So, the fourth bright fringe of wavelength 520 nm coincides with 5th bright fringe of wavelength 650 nm.