Question

A beam of microwaves whose wavelength is $0.052 \mathrm{m}$ is coming towards a rectangular aperture of width $0.35 \mathrm{m}$. Resultant diffraction pattern is observed on a wall at $8.0 \mathrm{m}$ distance from aperture. The distance between first and second order outer fringes will be:
(A) 1.3 m
(B) 1.8 m
(C) 1.19 m
(D) 2.5 m

Answer 1

Hint: We know that wavelength is the distance between identical points (adjacent crests) in the adjacent cycles of a waveform signal propagated in space or along a wire. In wireless systems, this length is usually specified in meters (m), centimetres (cm) or millimetres (mm). The range of wavelengths sufficient to provide a description of all possible waves in a crystalline medium corresponds to the wave vectors confined to the Brillouin zone. This indeterminacy in wavelength in solids is important in the analysis of wave phenomena such as energy bands and lattice vibrations. The wavelength range covers from 400 nm to 700 nm the range of the visual spectrum where the human visual system is most sensitive.

Complete step by step answer
We know that a high point is called a maximum (plural maxima). A low point is called a minimum (plural minima). The general word for maximum or minimum is extremum (plural extrema). We say local maximum (or minimum) when there may be higher (or lower) points elsewhere but not nearby.
We can continue with an odd multiple of a half wavelength, minimum oscillation (interference minimum). At the points where a crest and a trough meet together, both waves cancel out each other (interference minima) and where two crests or two troughs meet together, they reinforce each other (interference maxima).
Let us write the values that are mentioned in the question.
$\lambda=0.052 \mathrm{m}, \mathrm{a}=0.36 \mathrm{m}, \mathrm{D}=80 \mathrm{m}$
If first maxima is at distance X, from central maxima, then $\sin 0=\dfrac{3 \lambda}{2 \mathrm{a}}=\dfrac{\mathrm{X}_{1}}{\mathrm{D}} \mathrm{X}_{1}=\dfrac{3 \lambda \mathrm{D}}{2 \mathrm{a}}$
In the same way for second maxima $\mathrm{X}_{2}=\dfrac{5 \lambda \mathrm{D}}{2 \mathrm{a}}$
Distance between first and second maxima is given as:
$\Delta \mathrm{X}=\mathrm{X}_{2}-\mathrm{X}_{1}=\dfrac{5 \lambda \mathrm{D}}{2 \mathrm{a}}-\dfrac{3 \lambda \mathrm{D}}{2 \mathrm{a}}=\dfrac{2 \lambda \mathrm{D}}{2 \mathrm{a}}$
or $\Delta \mathrm{X}=\dfrac{\lambda \mathrm{D}}{\mathrm{a}}=\dfrac{0.052 \times 8}{0.35}$
$=1.1885 \mathrm{m}$
$=1.19 \mathrm{m}$

Hence option C is correct.

Note: We can conclude that a mechanical wave is a disturbance that is created by a vibrating object and subsequently travels through a medium from one location to another, transporting energy as it moves. This type of wave pattern that is seen traveling through a medium is sometimes referred to as a traveling wave. In physics a wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. Wave motion transfers energy from one point to another, often with no permanent displacement of the particles of the medium —that is, with little or no associated mass transport. Even though the wave speed is calculated by multiplying wavelength by frequency, an alteration in wavelength does not affect wave speed. Rather, an alteration in wavelength affects the frequency in an inverse manner. A doubling of the wavelength results in a halving of the frequency; yet the wave speed is not changed.