[2081 GIE ‘A’] In Young’s slit experiment, the separation of the first to fifth fringes is 2.5 mm when the wavelength used is 620 nm. The distance from the slits to the screen is 80 cm. Calculate the separation of two slits. [3] Ans: [latex]7.9\times 10^{-4}[/latex] m Solution Separation of first and 5th bright fringe (y) = 2.5 mm = 2.5 x 10-3 m Here, n = 4 So, [latex]\beta = \frac{y}{n}[/latex]
= [latex]\frac{2.5\times 10^{-3}}{4} = 6.25\times 10^{-4}[/latex] m Wavelength of light used ([latex]\lambda[/latex]) = 620 nm = [latex]620\times 10^{-9}[/latex] m = 6.2 x 10-7 m Distance from the slit to the screen (D) = 80 cm = 0.8 m Separation of the slits (d) =? We know that, [latex]\beta = \frac{\lambda D}{d}[/latex] Or, d = [latex]\frac{\lambda D}{\beta}[/latex] = [latex]\frac{(6.2\times 10^{-7})\times 0.8}{(6.25\times 10^{-4})}[/latex] = 7.936 x 10-4 m.
