$A$ slit of width $d$ is illuminated by white light. The first minimum for red light $(\lambda = 6500\,\mathring{A})$ will fall at $\theta = 30^\circ$ when $d$ will be

$A$ parallel beam of light of wavelength $\lambda$ is incident normally on a narrow slit. $A$ diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern,the phase difference between the rays coming from the two edges of the slit is:

In deriving the single slit diffraction pattern,it is stated that the intensity is zero at angles of $\theta = n\lambda / a$,where $a$ is the slit width. Justify this by suitably dividing the slit to bring out the cancellation.

How will the diffraction pattern change when yellow light is replaced by blue light? The fringe will be :-

In a single slit diffraction pattern,identify the incorrect statement from the following.

$A$ slit of width $a$ is illuminated by white light. For red light $(\lambda = 6500 \; \mathring{A})$,the first minima is obtained at $\theta = 30^\circ$. Then the value of $a$ will be