Consider the following Diagram :

By Huygen's Principle, we know every point on a Wavefront acts as a Secondary point source and emits light (wavelets in all direction).

Let us consider a Wavefront at a distance r , greater than the Opaque object.

By normal Ray Diagram, it is evident that the area beyond the Opaque object must be a Shadow. But as shown in the diagram, will not the Secondary Wavelets illuminate the shadow region too? So how are Shadows formed without contradicting Huygen's Principle?

• Google Huygens' Principle. – Hot Licks Mar 20 '18 at 22:24
• And also obligatory here is to read the story of the amusing and curious incident of the Poisson Spot. – WetSavannaAnimal Mar 20 '18 at 23:30
• Yes, no perfect 100% shadows exist. But think more: first employ an 'opaque object' which is less than one wavelength across. Then repeat, but use an object which is 100,000 wavelengths across. Also, google terms: Fresnel diffraction, Fraunhofer diffraction – wbeaty Mar 21 '18 at 1:10

Light does travel into the shadow area by way of diffraction. You can see it in the double slit or any multiple slit experiment as well as single edged.

• But in that case,as far as I understand, the slit width is comparable to Wavelength of light. But here the dimensions of the Opaque object is much larger than it – Madhuchhanda Mandal Mar 22 '18 at 11:50
• You don’t need a slit. Light diffracts around single edges also. – Bill Alsept Mar 22 '18 at 14:36

When a ray of light hits a surface it looses some energy, which depends on the surface material. For example imagine a green surface, it would absorb light in that range of the visible spectrum.

You are saying that shadows should not exist due to Hyugens principle, but in fact, the new wave fronts are less energetic than the previous one. After many collisions they would have much less frequency (energy) in the visible spectrum, so if it gets behind you (to your shadow), you would see no reflection.

• Really? So you mean Radio wave and Micro wave gets generated due to multiple reflections of visible light? – Madhuchhanda Mandal Mar 22 '18 at 11:48

You asked a very reasonable question. And the answer lays in the nature of light.

Any emission of light happens through the exciting of subatomic particles, mostly electrons. During the relaxation they emit photons. These photons are summarized called electromagnetic radiation. Due to the interpretation of slit experiments, were the pattern behind the slits (as well as behind any sharp edge) has a swelling intensity, from which is concluded about the wave nature of light. But again please consider that the fringes on the observation screen are composed of the same photons which at earlier time were emitted from electrons.

Now to verify that light gets dissipated in the same way, a water wave do, it is enough to install behind a long wall a photometer and to count the photons which arrive from a light spot near an edge of the wall. As far as I know, such a simple experiment was not carried out until now. And my prediction is, that in a otherwise dark room no one photon from the spot will be registered by the photometer (photon multiplier). My prediction is based on the photons nature of light and the manner, in which light gets influenced from sharpe edges.