In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

Two slits separated by $0.5 \,mm$ are illuminated by light of wavelength $500 \,nm$. The screen is at a distance of $120 \,cm$ from the slits. The phase difference between the interfering waves at a point $3 \,mm$ on the screen from the central bright fringe is ........... .

$A$ 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 a Young's double-slit experiment,interference is obtained using electron waves. The electrons are produced by an electron gun. To increase the fringe width,one must:

In a Young's double slit experiment,green light is incident on the two slits. The interference pattern is observed on a screen. Which of the following changes would cause the observed fringes to be more closely spaced?

Young's double-slit experiment is performed using green light,red light,and blue light,one at a time. The fringe widths recorded are $\beta_G, \beta_R,$ and $\beta_B$ respectively. Then: