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Question Number 95341 by AshrafNejem last updated on 24/May/20

Q) A light falls on two slits 0.15 mm apart.An interference pattern is produced on a screen 60 cm from the slits, if the distance between the second and the fifth bright bands (frings) is 0.7 cm. Calculate the avelength of the used light. solution: Δy_(n=2→n=5) =0.7⇒Δy=0.7/3=0.233×10^(−2) m Δy=((sλ)/d) ⇒λ=((dΔy)/s)=((0.15×10^(−3) ×0.233×10^(−2) )/(60×10^(−2) )) =5.83×10^(−7) =583 nm

$$\left.\mathrm{Q}\right)\:\mathrm{A}\:\mathrm{light}\:\mathrm{falls}\:\mathrm{on}\:\mathrm{two}\:\mathrm{slits}\:\mathrm{0}.\mathrm{15}\:{mm}\:\mathrm{apart}.\mathrm{An}\:\mathrm{interference}\:\mathrm{pattern} \\ $$$$\mathrm{is}\:\mathrm{produced}\:\mathrm{on}\:\mathrm{a}\:\mathrm{screen}\:\mathrm{60}\:{cm}\:\mathrm{from}\:\mathrm{the}\:\mathrm{slits},\:\mathrm{if}\: \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{second}\:\mathrm{and}\:\mathrm{the}\:\mathrm{fifth}\: \\ $$$$\mathrm{bright}\:\mathrm{bands}\:\left(\mathrm{frings}\right)\:\mathrm{is}\:\mathrm{0}.\mathrm{7}\:{cm}.\:\mathrm{Calculate}\:\mathrm{the}\: \\ $$$$\mathrm{avelength}\:\mathrm{of}\:\mathrm{the}\:\mathrm{used}\:\mathrm{light}. \\ $$$$\boldsymbol{{solution}}: \\ $$$$\Delta{y}_{{n}=\mathrm{2}\rightarrow\mathrm{n}=\mathrm{5}} =\mathrm{0}.\mathrm{7}\Rightarrow\Delta{y}=\mathrm{0}.\mathrm{7}/\mathrm{3}=\mathrm{0}.\mathrm{233}×\mathrm{10}^{−\mathrm{2}} \:{m} \\ $$$$\Delta{y}=\frac{{s}\lambda}{{d}}\:\Rightarrow\lambda=\frac{{d}\Delta{y}}{{s}}=\frac{\mathrm{0}.\mathrm{15}×\mathrm{10}^{−\mathrm{3}} ×\mathrm{0}.\mathrm{233}×\mathrm{10}^{−\mathrm{2}} }{\mathrm{60}×\mathrm{10}^{−\mathrm{2}} } \\ $$$$=\mathrm{5}.\mathrm{83}×\mathrm{10}^{−\mathrm{7}} =\mathrm{583}\:{nm} \\ $$

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