$A$ Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is

$A$ 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,if the distance between the slits is twice the wavelength,the number of possible interference maxima is:

$A$ beam of electrons is used in $Y.D.S.E.$ The slit width is $d$. When the velocity of the electrons is increased,then:

Minimum thickness of a mica sheet having $\mu = 3/2$ which should be placed in front of one of the slits in $YDSE$ is required to reduce the intensity at the centre of the screen to half of the maximum intensity is-

In a double slit experiment,the angular width of a fringe is found to be $0.2^{\circ}$ on a screen placed $1\, m$ away. The wavelength of light used is $600\, nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water (in $^{\circ}$)? Take the refractive index of water to be $4/3$.