Two parallel slits $0.6 \,mm$ apart are illuminated by a light source of wavelength $6000 \,\mathring{A}$. The distance between two consecutive dark fringes on a screen $1 \,m$ away from the slits is:

The fringe width in a $YDSE$ pattern is $2.4 \times 10^{-4} \, m$ when red light of wavelength $6400 \, \mathring{A}$ is used. How much will it change if blue light of wavelength $4000 \, \mathring{A}$ is used?

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:

The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width,the ratio of minimum to maximum intensity in the interference pattern is $x: 4$ where $x$ is ..... .

In Young's double slit interference experiment,the slit widths are in the ratio $1 : 25$. Then the ratio of intensity at the maxima and minima in the interference pattern is

In Young's double slit experiment,when the wavelength used is $6000 \ \mathring{A}$ and the screen is $40 \ cm$ from the slits,the fringes are $0.012 \ cm$ wide. What is the distance between the slits in $cm$?