Question

At a certain point on a screen in a YDSE apparatus, the path difference for the two interfering waves is ${\left(\frac{1}{3}\right)}^{rd}$ of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B

$\frac{1}{4}$

$\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$

$\Delta \varphi =\frac{2\pi }{3}$

$I=\left(I_0\right)co{s}^2{60}^{\circ }$

$\left(\frac{I}{I_0}\right)=\frac{1}{4}$