In Young's experiment,the separation between the $5^{\text{th}}$ maxima and the $3^{\text{rd}}$ minima is how many times the fringe width?

The distance between two slits is $3 \ mm$ and the screen is placed at $2 \ m$ distance. When blue-green light of wavelength $500 \ nm$ is used, then the distance between two consecutive fringes will be (in $mm$)?

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

In a Young's double-slit experiment with light of wavelength $\lambda$,the separation of slits is $d$ and the distance of the screen is $D$ such that $D >> d >> \lambda$. If the fringe width is $\beta$,the distance from the point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

In a $YDSE$,bichromatic light of wavelengths $\lambda_1 = 400\, nm$ and $\lambda_2 = 560\, nm$ is used. The distance between the slits is $d = 0.1\, mm$ and the distance between the plane of the slits and the screen is $D = 1\, m$. The minimum distance between two successive regions of complete darkness is.....$mm$.

In Young's double-slit experiment,the value of $\lambda = 500\, nm$. The value of $d = 1\, mm$ and $D = 1\, m$. Then the minimum distance from the central maximum for which the intensity is half the maximum intensity will be: