In Young's double slit experiment, with a source of light having wavelength $$6300 \mathop A\limits^o$$, the first maxima will occur when the

In Young's double slit experiment, the intensity at a point where the path difference is $$\frac{\lambda}{4}$$ [ $$\lambda$$ is wavelength of light used] is '$$\mathrm{I}$$'. If '$$\mathrm{I}_0$$' is the maximum intensity then $$\frac{\mathrm{I}}{\mathrm{I}_0}$$ is equal to $$\left[\cos \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]$$

In Young's double slit experiment, the '$$\mathrm{n^{th}}$$' maximum of wavelength '$$\lambda_1$$' is at a distance '$$\mathrm{y_1}$$' from the central maximum. When the wavelength of the source is changed to '$$\lambda_2$$', $$\left(\frac{\mathrm{n}}{2}\right)^{\text {th }}$$ maximum is at a distance of '$$\mathrm{y_2}$$' from its central maximum. The ratio $$\frac{y_1}{y_2}$$ is

Light of wavelength '$$\lambda$$' is incident on a single slit of width 'a' and the distance between slit and screen is 'D'. In diffraction pattern, if slit width is equal to the width of the central maximum then $$\mathrm{D}=$$