Question
Answer the following question:
Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.

Answer

The size of the obstacle or aperture should be comparable to the wavelength for diffraction of waves by obstacles, through a large angle.
This follows from $\text{sin}\theta=\frac{\lambda}{\text{a}}.$
For light waves,
Wavelength, $\lambda=10^{-7} \text{m and},$
Size of the wall, a = 10 m
$\therefore\ \text{sin}\theta=\frac{10^{-7}}{10}=10^{-8}$
This implies, $\theta\ \rightarrow\ 0$
That is, light goes almost unbent and hence, the students are unable to see each other.
For sound waves,
Frequency, $\nu$ = 1000 Hz
$\text{i.e}\ \frac{\text{c}}{\nu}=\frac{330}{1000}=0.33\ \text{m}$
$\therefore\ \text{sin}\theta=\frac{\lambda}{\text{a}}=\frac{0.33}{10}=0.033\ \text{m}$
Here, $\theta$ has a definite value and the waves will bend around the partition. Hence, students can converse easily.

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