Wave Optics Test - 3

For diffraction , the width of the slit must be less than the wavelength of the incident ray.

Diffraction of light  takes place for all types of waves  transverse orlongitudinal .

This is because for maximum diffraction the slit width always equals to the wavelength of light.

Beta(central Maxima) = 2 * lambda * D(distance of the screen) / d(distance between slits)
beta = 2 m, wavelength = 500nm = 5 ×10^-7 m &d = 0.1mm = 1 ×10^-4 m
►2 = 2 ×5 ×10^-7  * D/10^-4
►1 = 5 ×10^-3 D  
►D = 1000/5
► D = 200m

• Yes, sound waves are longitudinal, meaning that they oscillate parallel to the direction of their motion. So diffraction takes place in the sound waves.
• Since there is no component of a sound wave 's oscillation that is perpendicular to its motion, sound waves cannot be polarized .

The First minimum is located at the place where the central maximum ends. And the first minimum spreads further upto half the fringe width.

• When a source of white light is used instead of a monochromatic source, the diffracted image of the slit gets dispersed into constituent colours of white light.
• The central maximum will be white and on either side of the central maximum, there will be coloured fringes.

Actually, diffraction occurs all the time. But when the aperture/wavelength ratio is large , the spacing between fringes becomes too small, and the diffractive effects areunnoticeable .

Fringe width is inversely proportional to the distance between slits so it will be  doubled.

• Diffraction occurs with all waves, including sound waves, water waves, and electromagnetic waves such as visible light, X-rays and radio waves. For the diffraction of wave, an obstacle or aperture of the size of the wavelength of the wave is needed.
• Since the wavelength of the radio waves is large and comparable with the size of the obstacles around which they bend. So it shows maximum diffraction .