MHTCET 10 May 2024 Shift 2 Paper

Concept:In Young's double slit experiment, the positions of bright and dark fringes are given by standard formulas using slit separation, screen distance, and wavelength.Explanation:Given: slit separation $d = 0.6 \times 10^{-3}$ m and screen distance $D = 1.2$ m.The distance between the tenth bright fringe and the third dark fringe on the same side is $8.85 \times 10^{-3}$ m.Position of the $n$-th bright fringe is $y_n = \frac{n \lambda D}{d}$.For the tenth bright fringe: $y_{10} = \frac{10 \lambda (1.2)}{0.6 \times 10^{-3}} = 20 \times 10^{3} \lambda$.Position of the $m$-th dark fringe is $y'_m = \frac{(2m-1) \lambda D}{2d}$.For the third dark fringe: $y'_3 = \frac{5 \lambda (1.2)}{2 \times 0.6 \times 10^{-3}} = 5 \times 10^{3} \lambda$.Since both fringes are on the same side, their separation is $y_{10} - y'_3 = 8.85 \times 10^{-3}$.Substituting the expressions: $(20 \times 10^{3} - 5 \times 10^{3}) \lambda = 15 \times 10^{3} \lambda = 8.85 \times 10^{-3}$.Solving for $\lambda$: $\lambda = \frac{8.85 \times 10^{-3}}{15 \times 10^{3}} = 5.9 \times 10^{-7}$ m.Converting to Angstroms: $\lambda = 5900$ Å.Answer: The wavelength of light used is $5900$ Å, corresponding to option C.