In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is x units, $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{\lambda}{4}$ will be $\left(\cos 2 \pi=1, \cos \frac{\pi}{2}=0\right)$

In double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in interference pattern

In biprism experiment, the fringe width is 0.6 mm . The distance between $6^{\text {th }}$ dark fringe and $8^{\text {th }}$ bright fringe on the same side of central bright fringe is

In Young's double slit experiment, 'I' is the minimum intensity and ' $I_1$ ' is the intensity at a point where the path difference is $\frac{\lambda}{4}$ where ' $\lambda$ ' is the wavelength of light used. The ratio $I_1 \mathrm{I}_1$ is (Intensities of the two interfering waves are same) $\left(\cos 0^{\circ}=1, \cos 90^{\circ}=0\right)$