In Young's double slit experiment the separation between the slits is doubled without changing other setting of the experiment to obtain same fringe width, the distance 'D' of the screen from slit should be made

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

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