The distance for which ray optics becomes a good approximation for an aperture of $0.3 \ cm$ and a light of wavelength $6000 \ Å$ is (in $m$)

In a Fraunhofer diffraction,light of wavelength $\lambda$ is incident on a slit of width $d$. The diffraction pattern is observed on a screen placed at a distance $D$. If the linear width of the central maximum is equal to two times the width of the slit,then the value of $D$ is:

$A$ converging lens of a telescope with a diameter of $5 \text{ cm}$ has a focal length of $25 \text{ cm}$. In the focal plane of the lens, the distance between the centres of the Fraunhofer diffraction pattern is (wavelength of light used = $5000 \text{ Å}$, Rayleigh's criterion is satisfied).

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 $4500 \ Å$,then the distance between the first minimum on either side of the central maximum is [ $\theta$ is small and measured in radian.] (in $mm$)

The Fraunhofer spectrum is a

$A$ student is jogging on a straight path with a speed of $5.4 \,km/h$. Perpendicular to the path is a pipe with its opening $8 \,m$ from the road (see figure). The diameter of the pipe is $0.45 \,m$. At the other end of the pipe is a speaker emitting sound of $1280 \,Hz$ towards the opening of the pipe. As the student passes in front of the pipe,she hears the speaker sound for $T$ seconds. $T$ is in the range (Take speed of sound $= 320 \,m/s$):