Is Huygen's principle valid for longitudinal sound waves?
When we are considering a point source of sound wave. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of wavefront is in accordance with Huygen's principle.
So, Huygen's principle is valid for longitudinal sound waves also.
Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?
Consider the ray diagram shown below
The point image $I_1$, due to $L_1$ is at the focal point. Now, due to the converging lense $L_2$, let final image formed is $I$ which is point image, hence the wavefront for this image will be of spherical symmetry.
What is the shape of the wavefront on earth for sunlight?
We know that the sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity.
Due to the large distance the radius of wavefront can be considered as large (infinity) and hence, wavefront is almost plane.
Why is the diffraction of sound waves more evident in daily experience than that of light wave?
As we know that the frequencies of sound waves lie between 20 Hz to 20 kHz so that their wavelength ranges between 15 m to 15 mm . The diffraction occur if the wavelength of waves is nearly equal to slit width. As the wavelength of light waves is $7000 \times 10^{-10} \mathrm{~m}$ to $4000 \times 10^{-10} \mathrm{~m}$. The slit width is very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.
The human eye has an approximate angular resolution of $\phi=5.8 \times 10^{-4}$ rad and a typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch $=2.54 \mathrm{~cm}$ ). At what minimal distance $z$ should a printed page be held so that one does not see the individual dots.
Given, angular resolution of human eye, $\phi=5.8 \times 10^{-4} \mathrm{rad}$.
and printer prints 300 dots per inch.
The linear distance between two dots is $l=\frac{2.54}{300} \mathrm{~cm}=0.84 \times 10^{-2} \mathrm{~cm}$.
At a distance of $z \mathrm{~cm}$, this subtends an angle, $\phi=\frac{l}{z}$
$$\therefore\quad z=\frac{l}{\phi}=\frac{0.84 \times 10^{-2} \mathrm{~cm}}{5.8 \times 10^{-4}}=14.5 \mathrm{~cm}$$