In Young's double slit experiment,if the widths of the slits are in the ratio $4 : 9$,the ratio of the intensity at maxima to the intensity at minima will be

In a double-slit experiment performed in air, the angular width of a fringe is found to be $0.15^{\circ}$ on a screen placed $80 \ cm$ away. The wavelength of light used is $490 \ nm$. What is the angular width of the fringe if the entire apparatus is immersed in a medium of refractive index $\frac{5}{3}$ (in $^{\circ}$)?

In $Y.D.S.E.,$ the slits are $0.2 \, cm$ apart and are illuminated by yellow light $(\lambda = 600 \, nm).$ What would be the fringe width on a screen placed $1 \, m$ from the plane of the slits if the whole system is immersed in water of refractive index $\mu = \frac{4}{3}$ (in $, mm$)?

$A$ 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 a Young's double-slit experiment,the wavelength of light used is $5000 \, \mathring{A}$ and the fringe width obtained is $1 \, mm$. If the wavelength of light is changed to $6000 \, \mathring{A}$ while keeping the experimental setup unchanged,the new fringe width will be ........ $mm$.

Two slits are made $10 \ mm$ apart and the screen is placed $1.5 \ m$ away. What is the fringe separation when a wavelength of $7000 \ \mathring{A}$ is used (in $\mu m$)?