In YDSE wavelength of light used is 600nm . At a point intensity is 50 % of maximum value. What is the path difference at that point:-

Question

In YDSE wavelength of light used is 600nm . At a point intensity is 50 % of maximum value. What is the path difference at that point:- (A) 150nm (B) 200nm (C) 300nm (D) 600nm

Tardigrade Admin · Accepted Answer

I=Imax(cos)2((π /λ ) ⋅ Δ x) ⇒ (Imax/2)=Imax(cos)2((π /λ ) ⋅ Δ x) ⇒ (cos (π /λ ) ⋅ Δ x)=(1/√2)⇒ Δ x=(λ /4)

Q. In YDSE wavelength of light used is $600 nm$ . At a point intensity is $50%$ of maximum value. What is the path difference at that point:-

$150 nm$

$200 nm$

$300 nm$

$600 nm$

$I = I_{ma x} (cos)^{2} (\frac{π}{λ} \cdot Δ x)$
$\Rightarrow \frac{I _{ma x}}{2} = I_{ma x} (cos)^{2} (\frac{π}{λ} \cdot Δ x)$
$\Rightarrow (cos \frac{π}{λ} \cdot Δ x) = \frac{1}{2} \Rightarrow Δ x = \frac{λ}{4}$

Q. In YDSE wavelength of light used is 600nm . At a point intensity is 50% of maximum value. What is the path difference at that point:-

150nm

200nm

300nm

600nm