Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)

The intensity ratio of the two interfering beams of light is $\beta$. What is the value of $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$?

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

Two light beams of intensities in the ratio of $9: 4$ are allowed to interfere. The ratio of the intensity of maxima and minima will be

In a Young's double slit experiment,the intensity at a point is $\left(\frac{1}{4}\right)^{\text{th}}$ of the maximum intensity. The minimum distance of the point from the central maximum is . . . . . . . . $\mu m$.
(Given: $\lambda = 600 \ nm, d = 1.0 \ mm, D = 1.0 \ m$)

In a double slit experiment,the separation between the slits is $d = 0.25 \, cm$ and the distance of the screen $D = 100 \, cm$ from the slits. If the wavelength of light used is $\lambda = 6000 \mathring{A}$ and $I_0$ is the intensity of the central bright fringe,the intensity at a distance $x = 4 \times 10^{-5} \, m$ from the central maximum is