In Fraunhofer diffraction pattern, slit width is 0.3 mm and screen is at 1.5 m away from the lens. If wavelength of light used is $4500 \mathop {\rm{A}}\limits^{\rm{o}}$, then the distance between the first minimum on either side of the central maximum is [ $\theta$ is small and measured in radian.]

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is ' $I$ '. The intensity at a point where the path difference is $\lambda / 6$ is $\left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right] [\lambda=$ wavelength of light $][\cos \pi=-1]$

In Young's double slit experiment, the light of wavelength ' $\lambda$ ' is used. The intensity at a point on the screen is ' T ' where the path difference is $\lambda \frac{-}{4}$. If ' $\mathrm{I}_0$ ' denotes the maximum intensity then the ratio of ' $\mathrm{I}_0$ ' to ' I ' is $\left(\cos 45^{\circ}=1 / \sqrt{2}\right)$