A beam of light consisting of two wavelengths | vedclass

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(A) For the bright fringes of two different wavelengths to coincide at a distance $x$ from the central maximum,the condition is given by: $x = \frac{n

Vedclass · Accepted Answer

$0.156$

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(A) For the bright fringes of two different wavelengths to coincide at a distance $x$ from the central maximum,the condition is given by: $x = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$ This implies $n_1 \lambda_1 = n_2 \lambda_2$,where $n_1$ and $n_2$ are the orders of the bright fringes. Rearranging the terms,we get: $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{5200 \mathring{A}}{6500 \mathring{A}} = \frac{4}{5}$ For the least distance,we take the smallest integer values: $n_1 = 4$ and $n_2 = 5$. Now,substituting these values into the formula for $x$: $x = \frac{n_1 \lambda_1 D}{d} = \frac{4 	imes 6500 	imes 10^{-10} 	ext{ m} 	imes 1.2 	ext{ m}}{2 	imes 10^{-3} 	ext{ m}}$ $x = \frac{4 	imes 6500 	imes 1.2 	imes 10^{-10}}{2 	imes 10^{-3}} 	ext{ m} = 15600 	imes 10^{-7} 	ext{ m} = 1.56 	imes 10^{-3} 	ext{ m}$ $x = 0.156 	ext{ cm}$.

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