In Young's double slit experiment,the $y$-coordinates of the central maxima and the $10^{th}$ maxima are $2 \, cm$ and $5 \, cm$ respectively. When the $YDSE$ apparatus is immersed in a liquid of refractive index $\mu = 1.5$,what will be the corresponding $y$-coordinates?

In Young's double-slit experiment, one slit is wider than the other, such that the amplitude of the light wave from one slit is twice that of the other. If the maximum intensity is $I_m$, then the resultant intensity $I$ when they interfere with a phase difference of $\phi$ is:

In the Young's double slit experiment with sodium light,the slits are $0.589 \ m$ apart. The angular separation of the third maximum from the central maximum will be (given $\lambda = 589 \ nm$):

In Young's double slit experiment, for the $n^{th}$ dark fringe $(n = 1, 2, 3, \dots)$ the phase difference of the interfering waves in radian will be

$A$ light source,which emits two wavelengths $\lambda_1=400 \ nm$ and $\lambda_2=600 \ nm$,is used in a Young's double slit experiment. If recorded fringe widths for $\lambda_1$ and $\lambda_2$ are $\beta_1$ and $\beta_2$ and the number of fringes for them within a distance $y$ on one side of the central maximum are $m_1$ and $m_2$,respectively,then
$(A)$ $\beta_2 > \beta_1$
$(B)$ $m_1 > m_2$
$(C)$ From the central maximum,$3^{\text{rd}}$ maximum of $\lambda_2$ overlaps with $5^{\text{th}}$ minimum of $\lambda_1$
$(D)$ The angular separation of fringes for $\lambda_1$ is greater than $\lambda_2$

In Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $k$ units; $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\lambda/4$ will be: