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 1

Given in the question :-

The size of the obstacle should be comparable to wavelength for diffraction ,

Now ,

Sinθ = $\lambda /a$ ( where θ = diffraction angle)

a = size of the wall = 10 m.

for light , λ ≈ 10⁻⁷ m

sin θ = 10⁻⁷/ 10

sin θ = 10⁻⁸

⇒ θ → 0

Light is not at all bending that's why they can't see.

For sound waves, frequency ≈ 1000 Hz

λ = v/f

Sinθ = λ/a

sinθ = 0.033

⇒ θ = 1° 53'

Here θ has a definite non- zero value .

Sound waves will bend around partition and the students will be able to converse easily.

Hope it helps :-)