In Young's double slit experiment,the distance between the sources is $1 \ mm$ and the distance between the screen and the source is $1 \ m$. If the fringe width on the screen is $0.06 \ cm$,then $\lambda = \dots \mathring{A}$.

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 $\mu m$)?

In Young's double-slit experiment,the value of $\lambda = 500\, nm$. The value of $d = 1\, mm$ and $D = 1\, m$. Then the minimum distance from the central maximum for which the intensity is half the maximum intensity will be:

Two slits at a distance of $1\, mm$ are illuminated by light of wavelength $6.5 \times 10^{-7}\, m$. The interference fringes are observed on a screen placed at a distance of $1\, m$. The distance between the third dark fringe and the fifth bright fringe will be......$mm$.

In Young's double-slit experiment,using light of wavelength $\lambda = 5890 \ \mathring{A}$,the angular width of the fringe is $0.20^\circ$. To increase the angular width by $10\%$,the wavelength must be increased by .......... $\mathring{A}$.

In a Young's double slit experiment,the slit separation $d$ is $0.3 \text{ mm}$ and the screen distance $D$ is $1 \text{ m}$. $A$ parallel beam of light of wavelength $600 \text{ nm}$ is incident on the slits at angle $\alpha$ as shown in the figure. On the screen,the point $O$ is equidistant from the slits and the distance $PO$ is $11.0 \text{ mm}$. Which of the following statement$(s)$ is/are correct?