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 $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\angle S_1 C S_2 = \theta$ and $\lambda$ is the wavelength,the fringe width is

The wavelength of light $500 \, nm$ is used in a Young's double slit experiment. The distance between the slits and the screen is $100 \, cm$ and the slits are separated by $1 \, mm$. Find the distance between the fifth $(5^{th})$ and third $(3^{rd})$ bright fringes. (in $mm$)

$A$ student is asked to measure the wavelength of monochromatic light. He sets up the apparatus as shown in the figure. $S_1, S_2, S_3$ are narrow parallel slits,$L$ is a sodium lamp,and $M$ is a microscope eyepiece. The student fails to observe interference fringes. Your first advice to him will be:

In Young's double-slit experiment,if $L$ is the distance between the slits and the screen upon which the interference pattern is observed,$x$ is the average distance between adjacent fringes,and $d$ is the slit separation,then the wavelength of light is given by:

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?