• # question_answer In YDSE screen is kept 0.8 m from the slits. The coherent sources are 0.016 cm apart & fringes are observed on the screen. It is found that with a certain monochromatic source of light, the fourth bright fringe is situated at a distance of 1.06 cm from the central fringe. The wavelength of light used is A)  480nm                                 B)  530 nm   C)  650 nm                                D)  580 nm

For maxima, $\Delta x=n\lambda$$\Rightarrow$$\frac{dy}{D}=n\lambda$ For fourth maxima, n = 4 $\frac{dy}{D}=4\lambda \Rightarrow \lambda =\frac{dy}{4D}=\frac{0.016\times {{10}^{-2}}\times 1.06\times {{10}^{-2}}}{4\times 0.8}$ $\Rightarrow \lambda =530nm$