Young's experiment demonstrates that . . . . . .

In a Young's double-slit experiment,the separation of the two slits is doubled. To keep the same spacing of fringes,the distance $D$ of the screen from the slits should be made

In a Young's double slit experiment,the central point on the 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 a double-slit experiment using light of wavelength $600 \; nm$,the angular width of a fringe formed on a distant screen is $0.1^{\circ}$. What is the spacing between the two slits?

In $Y.D.S.E$, two slits are made $1 \, mm$ apart and the screen is placed $1 \, m$ away. The fringe separation when light of wavelength $500 \, nm$ is used is: