In $YDSE$ experiment,when two light rays form the third minima,they have:

In a Young's double slit experiment,the distance between the two identical slits is $6.1$ times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is

In a Young's double slit experiment,the intensities at two points,for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits,then $\frac{I_1 + I_2}{I_0} = \dots$

Assertion: In Young's double slit experiment,the two slits are at a distance $d$ apart. An interference pattern is observed on a screen at a distance $D$ from the slits. At a point on the screen directly opposite to one of the slits,a dark fringe is observed. Then,the wavelength of the wave is proportional to the square of the distance between the two slits.
Reason: For a dark fringe,the intensity is zero.

In Young's double slit experiment,the distance between slits and the screen is $1.0\,m$ and monochromatic light of $600\,nm$ is being used. $A$ person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied,the interference pattern disappears for a particular distance $d_0$ between the slits. If the angular resolution of the eye is $\frac{1}{60}^o,$ the value of $d_0$ is close to......$mm$

Light consisting of plane waves of wavelengths $\lambda_1 = 8 \times 10^{-5} \ cm$ and $\lambda_2 = 6 \times 10^{-5} \ cm$ generates an interference pattern in Young's double-slit experiment. If $n_1$ denotes the $n_1^{\text{th}}$ dark fringe due to light of wavelength $\lambda_1$ which coincides with the $n_2^{\text{th}}$ bright fringe due to light of wavelength $\lambda_2$, then: