In a Young's double slit experiment,the wavel | vedclass

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(B) The position of the $n^{	ext{th}}$ bright fringe from the central maximum is given by $y_n = \frac{n \lambda D}{d}$.For the $n^{	ext{th}}$ red b

Vedclass · Accepted Answer

$2$

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(B) The position of the $n^{	ext{th}}$ bright fringe from the central maximum is given by $y_n = \frac{n \lambda D}{d}$.For the $n^{	ext{th}}$ red bright band: $y_{R,n} = \frac{n \lambda_R D}{d}$.For the $(n+1)^{	ext{th}}$ blue bright band: $y_{B,n+1} = \frac{(n+1) \lambda_B D}{d}$.Given that these two bands coincide,we have $y_{R,n} = y_{B,n+1}$.Therefore,$\frac{n \lambda_R D}{d} = \frac{(n+1) \lambda_B D}{d}$.Canceling $D/d$ from both sides,we get $n \lambda_R = (n+1) \lambda_B$.Substituting the given values: $n(7.8 	imes 10^{-5}) = (n+1)(5.2 	imes 10^{-5})$.$7.8n = 5.2n + 5.2$.$2.6n = 5.2$.$n = 2$.

In a Young's double slit experiment,the wavelength of red light is $7.8 \times 10^{-5} \,cm$ and that of blue light is $5.2 \times 10^{-5} \,cm$. The value of $n$ for which $(n+1)^{\text{th}}$ blue bright band coincides with $n^{\text{th}}$ bright red band is ..........

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