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(C) Given, $\lambda_1 = 8 	imes 10^{-5} \ cm$ and $\lambda_2 = 6 	imes 10^{-5} \ cm$.The position of the $n_1^{	ext{th}}$ dark fringe due to light

Vedclass · Accepted Answer

$n_1=1, n_2=2$

Vedclass · Answer

(C) Given, $\lambda_1 = 8 	imes 10^{-5} \ cm$ and $\lambda_2 = 6 	imes 10^{-5} \ cm$.The position of the $n_1^{	ext{th}}$ dark fringe due to light of wavelength $\lambda_1$ is given by $x_{n_1} = (2n_1 - 1) \frac{D \lambda_1}{2d}$.The position of the $n_2^{	ext{th}}$ bright fringe due to light of wavelength $\lambda_2$ is given by $x_{n_2} = \frac{n_2 D \lambda_2}{d}$.Since the fringes coincide, $x_{n_1} = x_{n_2}$:$(2n_1 - 1) \frac{D \lambda_1}{2d} = \frac{n_2 D \lambda_2}{d}$$(2n_1 - 1) \frac{\lambda_1}{2} = n_2 \lambda_2$$\frac{2n_1 - 1}{n_2} = \frac{2 \lambda_2}{\lambda_1} = \frac{2 	imes 6 	imes 10^{-5}}{8 	imes 10^{-5}} = \frac{12}{8} = \frac{3}{2}$Thus, $2(2n_1 - 1) = 3n_2$, which simplifies to $4n_1 - 2 = 3n_2$.Testing the options:For $n_1=1, n_2=2$: $4(1) - 2 = 2$ and $3(2) = 6$ (No).Wait, re-evaluating the dark fringe formula: $x = (2n_1 - 1) \frac{D \lambda_1}{2d}$. If $n_1=1$, $x = \frac{D \lambda_1}{2d}$.Using $\frac{2n_1-1}{n_2} = \frac{3}{2}$, if $n_1=2$, $\frac{4-1}{n_2} = \frac{3}{n_2} = \frac{3}{2} \implies n_2=2$.Checking option $C$: $n_1=1, n_2=2 \implies \frac{2(1)-1}{2} = 1/2 
eq 3/2$.Checking option $D$: $n_1=3, n_2=2 \implies \frac{2(3)-1}{2} = 5/2 
eq 3/2$.Actually, for $n_1=2, n_2=2$, the condition is satisfied. Given the options, let's re-check the calculation: $4n_1 - 3n_2 = 2$. If $n_1=2, n_2=2$, $8-6=2$. If $n_1=1, n_2=2/3$ (invalid). The correct pair is $n_1=2, n_2=2$.

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