Q.Why is the diffraction of sound waves more evident in daily experience than that of light wave?

Diffraction — the bending of a wave around obstacles or through openings — is appreciable only when the wavelength is comparable to the size of the obstacle/aperture ($\theta \sim \lambda/d$). Sound in air has wavelengths from about $1.7\ \text{cm}$ to $17\ \text{m}$, the same scale as doors, windows and furniture, so it bends around them noticeably. Visible light has wavelengths of only $400$–$700\ \text{nm}$, millions of times smaller than everyday objects, so its diffraction is imperceptible without a very fin …

1. The condition for noticeable diffraction. For any wave, the angular spread on passing an obstacle or aperture of size $d$ is of order

The bending is significant only when $\lambda$ is comparable to (or larger than) $d$; if $\lambda \ll d$ the wave travels almost in straight lines.

2. Wavelength of sound. With $v \approx 340\ \text{m/s}$ and audible frequencies $20\ \text{Hz}$–$20\ \text{kHz}$,

These lengths are the same order as doors, windows, pillars and people, so $\lambda/d \sim 1$ and sound diffracts strongly — you hear people around corners.

3. Wavelength of light. Visible light has $\lambda \approx 400$–$700\ \text{nm} = (4$–$7)\times10^{-7}\ \text{m}$. For a door of width $\sim 1\ \text{m}$, $\lambda/d \sim 10^{-7}$ — the bending is utterly negligible, so light casts sharp shadows and appears to travel in straight lines. …

Method: Comparing Diffraction Significance Across Waves

Use this whenever you must judge whether diffraction will be noticeable for a

given wave meeting an obstacle or aperture of a given size — not to calculate

an exact diffraction pattern, just to decide "is bending obvious here or not."

Steps

Step 1: Identify the two lengths that matter

Diffraction is a comparison between the wave's wavelength $\lambda$ and the

characteristic size $d$ of whatever it is passing (a slit, an edge, a door, an

obstacle). Write down both.

Step 2: Apply the significance criterion

This is the rough angular spread a wave acquires on passing an obstacle/aperture

of size $d$. The rule of thumb:

• If $\lambda \sim d$ or $\lambda > d$: diffraction is strong and easily observed.
• If $\lambda \ll d$: the wave travels essentially in straight lines; diffraction is present in principle but invisible without a very fine slit.

Step 3: Compute the wavelength(s) involved

For a mechanical wave (e.g. sound), get $\lambda$ from its speed and frequency

range: $\lambda = v/f$. For light, use the known visible range …