In Young’s double slit experiment, how many maxima can be seen on a screen (including the central maxima) if $d = \frac{5\lambda}{2}$ (where $\lambda$ is the wavelength of light and $d$ is the distance between the two slits)?

$A$ Young's double-slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is

In a standard $YDSE$ setup,two sources $S_1$ and $S_2$ of intensities $I_1$ and $I_2$ are placed in front of a screen (figure). The pattern of intensity distribution seen in the central portion is given by the graph. In this case,which of the following statements is true?

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ bright band and the $5^{\text{th}}$ bright band. The wavelength of the light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

In Young's double-slit experiment,the fringe width on the screen is $0.2 \, mm$. If the wavelength of the light used is increased by $10\%$ and the distance between the two slits $S_1$ and $S_2$ is also increased by $10\%$,the new fringe width will be ....... $mm$.

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: