Two slits are made $10 \ mm$ apart and the screen is placed $1.5 \ m$ away. What is the fringe separation when a wavelength of $7000 \ \mathring{A}$ is used (in $\mu m$)?

In Young's double slit experiment,the two slits are $0.6 \; mm$ apart. The interference pattern is observed on a screen at a distance of $80 \; cm$ from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light is $\dots \; nm$.

$A$ fringe width of a certain interference pattern is $\beta = 0.002 \text{ cm}$. What is the distance of the $5^{\text{th}}$ dark fringe from the center?

Two slits at a distance of $1\, mm$ are illuminated by light of wavelength $6.5 \times 10^{-7}\, m$. The interference fringes are observed on a screen placed at a distance of $1\, m$. The distance between the third dark fringe and the fifth bright fringe will be......$mm$.

In Young's double-slit experiment,monochromatic light is used to illuminate two slits $A$ and $B$. Interference fringes are obtained on a screen in front of the slits. Now,if a thick glass plate is placed in the path of light from one of the slits,then:

In Young's experiment, the intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources is $36 : 1$. The ratio of the amplitudes of the two individual sources will be