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

In a Young's double slit experiment,the intensities at two points,for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits,then $\frac{I_1 + I_2}{I_0} = \dots$

In a Young's double-slit experiment,light of wavelength $5890 \ \mathring A$ is used,and the angular fringe width on the screen is $0.20^\circ$. If the entire apparatus is immersed in water,find the new angular fringe width. (Refractive index of water $\mu = 4/3$) (in $^\circ$)

In Young's double slit experiment,the light emitted from the source has $\lambda = 6.5 \times 10^{-7} \, m$ and the distance between the two slits is $1 \, mm$. The distance between the screen and the slits is $1 \, m$. The distance between the third dark fringe and the fifth bright fringe will be ......... $mm$.

$A$ 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:

Electrons accelerated from rest by an electrostatic potential are collimated and sent through a Young's double slit experiment. The fringe width is $\omega$. If the accelerating potential is doubled,then the width is now close to ............. $\omega$.