$A$ beam of light consisting of wavelengths $650 \ nm$ and $550 \ nm$ illuminates the Young's double slits with separation of $2 \ mm$ such that the interference fringes are formed on a screen, placed at a distance of $1.2 \ m$ from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is . . . . . . $\times 10^{-5} \ m$.

$A$ beam of light consisting of two wavelengths $650 \, nm$ and $520 \, nm$ is used to illuminate the slits of a Young's double-slit experiment. The order of the bright fringe of the longer wavelength that coincides with a bright fringe of the shorter wavelength at the least distance from the central maximum is:

What happens to the fringe width in Young's double-slit experiment if it is performed in glycerine instead of air?

What is the effect on the interference fringes in a Young's double-slit experiment due to each of the following operations:
$(a)$ the screen is moved away from the plane of the slits;
$(b)$ the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength;
$(c)$ the separation between the two slits is increased;
$(d)$ the source slit is moved closer to the double-slit plane;
$(e)$ the width of the source slit is increased;
$(f)$ the monochromatic source is replaced by a source of white light?
(In each operation,take all parameters,other than the one specified,to remain unchanged.)

$A$ 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?

$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width