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Wave Optics — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsWave Optics3 Marks
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.
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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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