When I sit in a room I can hear voices coming from the adjacent room but the light in adjacent room does not enter my room i.e. sound waves travels through the wall but light waves can't. Why?
[Physics] Why can’t light pass through walls but sound can
acousticsvisible-lightwaves
Related Solutions
The problem is acoustic impedance matching at the interface between the air and the wall. There an analogue of Fresnel's equations dictates how much power goes into the reflected and transmitted waves, and not surprisingly the more dissimilar the materials are the more power goes into the reflected wave. You can evade this by putting the speaker directly on the wall, effectively changing the problem so that the impedance matching is now between the source and the wall.
For a slightly surprising take on acoustic impedance, see this WP article.
The sounds that you hear in the kitchen have a frequency range of (say) 50 - 500 Hz, with smatterings of higher frequencies (dishes banging) into the low kHz.
The speed of sound is approximately 300 m/s. This means that a 300 Hz wave form has a wavelength of about 1 m. This in turn means that when that (sound) wave bounces off an object that is locally "smooth compared to 1 m", it maintains coherence - the bits of sound bouncing off one part of the wall will remain in phase with bits of sound bouncing off another bit of wall.
Another way of looking at this: at any moment, you can consider every point of the wave front (sound, light) to be a series of point sources, all of which are in phase. As long as the dimension of openings etc. is small (comparable to the wavelength), these wave will spread out - this is diffraction. You can see a nice image at http://homepages.ius.edu/kforinas/S/pics/diffraction1.jpg
By contrast, visible light has wavelengths in the 400 - 700 nm range. Thus photons bouncing off one bit of the surface will have a random phase relationship relative to other photons. The net result is that these photons (waves) will interfere constructively in just one specific direction (like Young's diffraction experiment, they will interfere constructively when their phase difference is a multiple of $2\pi$. ) In practice this means that the photons end up scattered, losing their spatial coherence. That is a fancy way of saying "they get so scrambled that they no longer form an image".
Recognize that imaging requires that there is a 1:1 relationship between the origin of a photon (did it come from the pan, or from the wife), and the direction in which it arrives at your eye. The lens in the eye makes sure that all photons that arrived from one direction end up on the same point on your retina. As long as all these photons came from the same source, that means that source becomes a "point" in the image you see. When the directions of the photons get scrambled, all you end up seeing is an "average intensity" of the light. Think "shower door", but worse. [note the above is slightly simplified, assuming that the object you look at is at infinity. Closer up there is a very small divergence of the light reaching different parts of your eye - this requires you to "focus" with your lens. That doesn't affect the basic principle in play here.]
This is why you can tell whether the light is on in the kitchen at the end of the hall, but you can't see what is going on - unless the walls are made of mirrors (which are smooth on the scale needed for specular reflection)
As an aside - sound can also travel "through the wall"; the pressure waves will cause slight motion of the wall which in turn causes waves on the other side (although much reduced in amplitude because of the acoustic mismatch between the air and the wall; this is where the "glass against the wall" so beloved in 50's movies can help). In principle, electromagnetic waves can also travel through walls (which is why your radio works indoors) - but again, the short wavelength of light has two implications. One is, that the distance to cross is "many wavelengths" so that even a tiny extinction coefficient is sufficient to cause complete absorption; and the short wavelength means that "everything" in the wall acts as a scattering point, and the light doesn't stand a chance of making it through in measurable quantities.
Best Answer
Whether any form of wave can pass through an object depends on how strongly that wave is reflected, scattered or absorbed.
Sound waves are certainly reflected by a wall, otherwise you wouldn't hear an echo from it, but not all the sound is reflected so some travels into the wall. Whether the sound is scattered and/or absorbed in the wall depends on what the wall is made from. Remember that sound is a mechanical vibration. The sound hitting the wall makes the wall vibrate and the other side of the wall makes the air on the other side vibrate. A good solid wall won't disperse the vibrations too much, so you will get some sound through it. A wall filled with e.g. fibreglass insulation, will absorb the sound far more, so it will transmit less sound.
Light is also reflected by a wall, otherwise you wouldn't be able to see it. How much light is reflected depends on the wall: a white painted wall will reflect more of the light than a black painted one. However the dominant interaction with the wall is probably scattering. If the wall was made of glass then obviously the light would pass through it. A concrete wall is made from miniscule grains of calcium carbonate and aluminosilicates, and while these materials are transparent to visible light, reflections from all those grain boundaries scatter light strongly. If a concrete wall was very thin e.g. 0.1 mm you would still get some light transmission through it.
Response to Zeynel's comment:
Consider a microphone. Sound waves consist of to and fro oscillations of air molecules, and if you sit at a fixed point in space these to and fro oscillations create oscillating pressure changes. A microphone works because when the pressure is high is pushes the sensor in the microphone back, and when the pressure is low it pulls it forward. The end result is to make the sensor in the microphone oscillate in time with the wave, and in the microphone this movement is used to create an oscillating electric field.
A microphone is designed to be very sensitive to changes in air pressure, and a wall is not. Nevertheless, even a solid wall is elastic in the sense that it deforms when you push on it. So a wall will also move in response to the oscillating pressure created by a sound wave. It will move much, much less than a microphone sensor, but it will move. If the side of the wall facing the sound wave oscillates then obviously the of the wall will oscillate as well. This behaves like a loudspeaker, i.e. the opposite of a microphone, as the oscillating surface of the wall creates an oscillating pressure in the air next to it, and this creates a sound wave. That's how the sound gets through the wall.
Re your last comment "if we said some of the sound waves passes through the wall": you need to remember that a sound wave is not a thing. It's just the movement of something else. In air a sound wave is a movement of the atoms in air, and in a wall a sound wave is the movement of the atoms in the wall. It's true to say that the sound wave passes through the wall, but it's the vibration that moves through the wall not anything you could point to. In this respect sound is completely different to light, where in principle you could follow a photon as it moves between different media.