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 $\lambda/4$.

Two coherent sources of intensity ratio $\alpha$ interfere. The value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is

In Young's double slit experiment,monochromatic light of wavelength $5000 \ \mathring{A}$ is used. The slits are $1.0 \ mm$ apart and the screen is placed at $1.0 \ m$ away from the slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is . . . . . . $\times 10^{-6} \ m$.

In Young's double slit experiment,one of the slits is wider than the other,so that the amplitude of the light from one slit is double that of the other. If $I_m$ is the maximum intensity,the resultant intensity $I$ when they interfere at a phase difference $\phi$ is given by

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

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: