In Young's double-slit experiment,for a distance of $1.8 \lambda$ between the two slits,what is the maximum number of possible interference maxima? Here,$\lambda$ is the wavelength of light.

In Young's double-slit experiment,the intensity at a point on the screen is $I_0/4$. Find the angular position of this point.

In Young's double-slit experiment,if the amplitudes of the interfering waves are not equal,then . . . . .

The angular width of a fringe in a double slit experiment is found to be $0.2^{\circ}$ on a screen $1 \ m$ away. The wavelength of light used is $600 \ nm$. What is the change in the angular width of the fringe if the entire measurement system is immersed in water (in $^{\circ}$)? (Use refractive index of water as $\mu = \frac{4}{3}$)

In Young's double-slit experiment, one slit is wider than the other, such that the amplitude of the light wave from one slit is twice that of the other. If the maximum intensity is $I_m$, then the resultant intensity $I$ when they interfere with a phase difference of $\phi$ is:

The distance of the $5^{\text{th}}$ dark fringe from the center is $4 \, mm$. If $D = 2 \, m$ and $\lambda = 600 \, nm$,then the distance between the slits is (in $mm$):