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 $$\mathrm{D}=$$

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2m away from the lens. If wavelength of light used is 5000$$\mathop A\limits^o $$ then the distance between the first minimum on either side of the central maximum is ($$\theta$$ is small and measured in radian)

A graph is plotted between the fringe-width Z and the distance D between the slit and eye-piece, keeping other adjustment same. The correct graph is

(A)

(B)

(C)

(D)

The Brewster's angle for the glass-air interface is $(54.74)^{\circ}$. If a ray of light passing from air to glass strickes at an angle of incidence $45^{\circ}$, then the angle of refraction is

$$\left[\tan (54.74)^{\circ}=\sqrt{2}, \sin 45=\frac{1}{\sqrt{2}}\right]$$