The path difference between two interfering waves at a point on the screen is $\frac{\lambda}{8}$. The ratio of intensity at this point and that at the central fringe will be

The ratio of maximum to minimum intensity due to the superposition of two waves is $\frac{49}{9}$. Then the ratio of the intensities of the component waves is:

Two coherent light sources having intensity in the ratio $2x$ produce an interference pattern. Then the value of $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$ will be

In $YDSE$,the intensity of the central bright fringe is $8 \, mW/m^2$. What will be the intensity at a path difference of $\frac{\lambda}{6}$?

In a $YDSE$ setup,the intensity of two coherent beams differs by $1\%$. If one of the beams has intensity $I$,then the intensity of the minima is:

In Young's double slit experiment, the wavelength of light used is $\lambda$. The intensity on the screen at a point for path difference $\lambda/6$ is $X$. The intensity at the point for path difference $\lambda/3$ is (Given: $\cos 180^{\circ} = -1, \cos 30^{\circ} = \frac{\sqrt{3}}{2}$)