Two sources of light $$0.6 \mathrm{~mm}$$ apart and screen is placed at a distance of $$1.2 \mathrm{~m}$$ from them. A light of wavelength $$6000\,\mathop A\limits^o$$ used. Then the phase difference between the two light waves interfering on the screen at a point at a distance $$3 \mathrm{~mm}$$ from central bright band is

Light of wavelength ',$$\lambda$$' is incident on a slit of width '$$\mathrm{d}$$'. The resulting diffraction pattern is observed on a screen at a distance '$$D$$'. The linear width of the principal maximum is then equal to the width of the slit if $$D$$ equals

In Young's double slit experiment, the wavelength of light used is '$$\lambda$$'. The intensity at a point is '$$\mathrm{I}$$' where path difference is $$\left(\frac{\lambda}{4}\right)$$. If $$I_0$$ denotes the maximum intensity, then the ratio $$\left(\frac{\mathrm{I}}{\mathrm{I}_0}\right)$$ is

$$\left(\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right)$$

In Young's double slit experiment, the fringe width is $$2 \mathrm{~mm}$$. The separation between the $$13^{\text {th }}$$ bright fringe and the $$4^{\text {th }}$$ dark fringe from the centre of the screen on same side will be