Test: Wave Optics- 1 (March 4) - NEET MCQ

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 or longitudinal.

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^-7m & d = 0.1mm = 1×10^-4m
► 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 are unnoticeable.

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.