In Young's double-slit experiment,the angular width of the fringes formed on the screen is $\pi / 200$. If light of wavelength $4800 \, \mathring{A}$ is used,find the distance between the slits.

The fringe width in Young's double-slit experiment can be increased if we decrease

Two ideal slits $S_1$ and $S_2$ are at a distance $d$ apart and are illuminated by light of wavelength $\lambda$ passing through an ideal source slit $S$ placed on the line through $S_2$ as shown. The distance between the plane of the slits and the source slit is $D$. $A$ screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is

Sodium light $(\lambda = 6 \times 10^{-7} \ m)$ is used to produce an interference pattern. The observed fringe width is $0.12 \ mm$. The angle between the two wave trains is:

In Young's double-slit experiment,the width of one slit is twice that of the other. What is the ratio of the maximum to the minimum intensity?

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 $mm$)