In a Young's double-slit experiment,the intensity at a point on the screen where the path difference is $\lambda$ is $K$. Find the intensity at a point where the path difference is $\lambda/4$.

In Young's double slit experiment,the intensity of light coming from the first slit is double the intensity from the second slit. The ratio of the maximum intensity to the minimum intensity on the interference fringe pattern observed is

Waves emitted from two identical sources produce an intensity of $K$ units at a point on the screen. If the path difference between these two waves is $\lambda$,calculate the intensity at a point on the screen where the path difference is $\lambda/4$.

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 path difference between two interfering waves of equal intensities at a point on the screen is $\frac{\lambda}{4}$. The ratio of intensity at this point and that at the central fringe will be

In a Young's double slit experiment, the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits, then $\frac{I_1 + I_2}{I_0} = $ $(\cos 45^{\circ} = \frac{1}{\sqrt{2}}, \cos 60^{\circ} = \frac{1}{2})$