The YDSE apparatus is as shown in the figure. The condition for point $P$ to be a dark fringe is ( $l =$ wavelength of light waves)

(l_{1} - l_{3}) + (l_{2} - l_{4}) = n λ

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(l_{1} - l_{2}) + (l_{3} - l_{4}) = \frac{(2 n + 1) λ}{2}

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(l_{1} + l_{3}) + (l_{2} + l_{4}) = \frac{(2 n - 1) λ}{2}

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(l_{1} - l_{2}) + (l_{4} - l_{3}) = \frac{(2 n - 1) λ}{2}

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Solution

The correct option is D $(l_{1} - l_{2}) + (l_{3} - l_{4}) = \frac{(2 n + 1) λ}{2}$
Path difference between $S S_{1} P$ and $S S_{2} P$ is
$Δ = (l_{1} + l_{3}) - (l_{2} + l_{4}) = (n + \frac{1}{2}) λ$
$\Rightarrow Δ = (l_{1} - l_{2}) + (l_{3} - l_{4}) = (n + \frac{1}{2}) λ$

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In the figure, if $l_{1} | | l_{2}$ and $l_{3} | | l_{4}$ what is y in terms of x ?

There are four lines such that, line $l_{1} ∥ l_{2}$ , line $l_{3} ⊥ l_{1}$ and line $l_{4} ⊥ l_{3}$ then which of the following option(s) is/are true?

The representation of $l_{1} parallel to l_{2}$ and $l_{3} perpendicular to l_{2}$ is respectively,

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The YDSE apparatus is as shown in the figure. The condition for point P to be a dark fringe is (l=wavelength of light waves)

The correct option is D (l1−l2)+(l3−l4)=(2n+1)λ2Path difference between SS1P and SS2P is Δ=(l1+l3)−(l2+l4)=(n+12)λ⇒Δ=(l1−l2)+(l3−l4)=(n+12)λ

The YDSE apparatus is as shown in the figure. The condition for point $P$ to be a dark fringe is ( $l =$ wavelength of light waves)

The correct option is D $(l_{1} - l_{2}) + (l_{3} - l_{4}) = \frac{(2 n + 1) λ}{2}$
Path difference between $S S_{1} P$ and $S S_{2} P$ is
$Δ = (l_{1} + l_{3}) - (l_{2} + l_{4}) = (n + \frac{1}{2}) λ$
$\Rightarrow Δ = (l_{1} - l_{2}) + (l_{3} - l_{4}) = (n + \frac{1}{2}) λ$