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 $\lambda/4$.

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 . . . . . . $\times 10^{-6} \ m$.

Obtain the conditions for constructive interference and destructive interference.

Two waves have their amplitudes in the ratio $1 : 9$. The maximum and minimum intensities when they interfere are in the ratio

Consider an $YDSE$ that has different slit widths. As a result,the amplitudes of waves from the two slits are $A$ and $2A$,respectively. If $I_0$ is the maximum intensity of the interference pattern,then the intensity of the pattern at a point where the phase difference between the waves is $\phi$ is:

In a standard $YDSE$ setup,two coherent sources of light of intensity ratio $\beta$ produce an interference pattern. The value of $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$ is equal to (where $I_{\max}$ and $I_{\min}$ are the maximum and minimum intensities of the resultant wave).