Consider the YDSE arrangement shown in figure- If d - 10- then the position of the 8th maxima is

For maxima, dsinθ =nλ 8th maximum → n=8 ⇒sinθ =nλd=8λ10λ ∴sinθ=45⇒tanθ=43 Also, tanθ=yD ∴ y=4D3 ...

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Standard XII

Physics

Tension in a String

Consider the ...

Question

Consider the YDSE arrangement shown in figure. If $d = 10 λ$ , then the position of the 8th maxima is

y = \frac{D}{10}

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y = \frac{D}{3}

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y = \frac{4 D}{5}

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y = \frac{4 D}{3}

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Solution

The correct option is D $y = \frac{4 D}{3}$
For maxima, $d sin θ = n λ$
8th maximum $\to n = 8$
$\Rightarrow sin θ = \frac{n λ}{d} = \frac{8 λ}{10 λ}$
$∴ sin θ = \frac{4}{5} \Rightarrow tan θ = \frac{4}{3}$
Also, $tan θ = \frac{y}{D}$
$∴ y = \frac{4 D}{3}$

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Consider the YDSE arrangement shown in figure. If d=10λ, then the position of the 8th maxima is

The correct option is D y=4D3For maxima, dsinθ=nλ 8th maximum →n=8 ⇒sinθ=nλd=8λ10λ ∴sinθ=45⇒tanθ=43 Also, tanθ=yD ∴y=4D3

Consider the YDSE arrangement shown in figure. If $d = 10 λ$ , then the position of the 8th maxima is

The correct option is D $y = \frac{4 D}{3}$
For maxima, $d sin θ = n λ$
8th maximum $\to n = 8$
$\Rightarrow sin θ = \frac{n λ}{d} = \frac{8 λ}{10 λ}$
$∴ sin θ = \frac{4}{5} \Rightarrow tan θ = \frac{4}{3}$
Also, $tan θ = \frac{y}{D}$
$∴ y = \frac{4 D}{3}$