In the Young's double slit experiment using a monochromatic light of wavelength λ the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is

Question

In the Young's double slit experiment using a monochromatic light of wavelength λ the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is (A) (2n+1) (λ/2) (B) (2n+1) (λ/4) (C) (2n+1) (λ/8) (D) (2n+1) (λ/16)

Tardigrade Admin · Accepted Answer

(I max /2)=I m cos 2((φ/2)) ⇒ cos ((φ/2))=(1/√2) ⇒ (φ/2)=(π/4) ⇒ φ=(π/2)(2 n +1) ⇒ Δ x =(λ/2 π) φ=(λ/2 π) × (π/2)(2 n +1)=(λ/4)(2 n +1)

Q. In the Young's double slit experiment using a monochromatic light of wavelength $λ$ the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is

$(2 n + 1) \frac{λ}{2}$

$(2 n + 1) \frac{λ}{4}$

$(2 n + 1) \frac{λ}{8}$

$(2 n + 1) \frac{λ}{16}$

$\frac{I _{m a x}}{2} = I_{m} cos^{2} (\frac{ϕ}{2})$
$\Rightarrow cos (\frac{ϕ}{2}) = \frac{1}{2}$
$\Rightarrow \frac{ϕ}{2} = \frac{π}{4}$
$\Rightarrow ϕ = \frac{π}{2} (2 n + 1)$
$\Rightarrow Δ x = \frac{λ}{2 π} ϕ = \frac{λ}{2 π} \times \frac{π}{2} (2 n + 1) = \frac{λ}{4} (2 n + 1)$

Q. In the Young's double slit experiment using a monochromatic light of wavelength λ the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is

(2n+1)2λ​

(2n+1)4λ​

(2n+1)8λ​

(2n+1)16λ​