In an interference experiment,let $D$ be the distance between the slit and the eyepiece. When the distance between two virtual sources is changed from $d_{A}$ to $d_{B}$,the fringe width changes from $Z_{A}$ to $Z_{B}$. The ratio $Z_{A} / Z_{B}$ is:

In Young's double-slit experiment,the distance between the two slits is $1 \, mm$ and the distance between the slits and the screen is $1 \, m$. If the wavelength of the light used is $500 \, nm$,then the distance of the third minimum from the central maximum is ... $mm$.

In Young's double slit experiment,the separation $d$ between the slits is $2 \ mm$,the wavelength $\lambda$ of the light used is $5896 \ \mathring{A}$,and the distance $D$ between the screen and slits is $100 \ cm$. It is found that the angular width of the fringes is $0.20^\circ$. To increase the fringe angular width to $0.21^\circ$ (with the same $\lambda$ and $D$),the separation between the slits needs to be changed to ...... $mm$.

In a $YDSE$ experiment,if a slab whose refractive index can be varied is placed in front of one of the slits,then the variation of resultant intensity at the mid-point of the screen with $\mu$ will be best represented by $(\mu \geq 1)$. [Assume slits of equal width and there is no absorption by the slab]

Two slits,$4 \, mm$ apart,are illuminated by light of wavelength $6000 \, \mathring{A}$. What will be the fringe width on a screen placed $2 \, m$ from the slits in $mm$?

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?