The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

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:

$A$ 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 . . . . . . $\times 10^{-5} \ m$.

White light is used to illuminate two slits in a $YDSE$. The separation between the slits is $d$ and the screen is at a distance $D (D >> d)$ from the slits. At a point on the screen directly in front of one of the slits,which of the following wavelengths are missing?

In Young's double slit experiment,the seventh maximum with wavelength $\lambda_1$ is at a distance $d_1$ from the central maximum,and the same maximum with wavelength $\lambda_2$ is at a distance $d_2$. Then $d_1/d_2$ is:

The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{87}{2}\right) \lambda$. The band obtained at that point is