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(a) Diffraction is the bending or spreading of a wave round an obstacle or as it passes through an opening. (b) (i) Critical angle is the angle of incidence in the denser medium when an angle of refraction in the lens dense medium is 90^\circ. (ii) (\(\alpha\))Critical angle = 90^\circ -.44^\circ = 46^\circ (\(\beta\))n = \(\frac{1}{sin c} = \frac{1}{\sin 46^o}\) = 1.39 (c) (i) Closed Pipe 1st Overtone = 3f\(_o\) = 3 x 200 = 600Hz Open Pipe All harmonics are possible overtone = 2f\(_o\) 2f\(_o\) = 600 f\(_o\) = 300Hz (ii) Closed Pipe f\(_o\) = \(\frac{v}{4l}\) l = \(\frac{v}{4f_o}\) \(\frac{330}{4 \times 200}\) = 0.412m Open Pipe \(f_1\) = \(\frac{v}{l}\) 600 = \(\frac{300}{l}\) l = 0.55m or \(f_1\) = \(\frac{v}{2l}\) 300 = \(\frac{300}{2 \times l}\) l = \(\frac{330}{600}\) = 0.55m

Soldiers march disorderly while crossing a bridge to prevent resonance because when soldiers march in orderly manner across a bridge, they generate a rhythmic oscillation of wave on the bridge and at a certain point, would start oscillation to the same rhythm as that of the marching steps. This oscillation would reach a maximum peak when the bridge can no longer sustain its own strength and hence collapses. So for this reasons, soldier are ordered to march disorderly while crossing a bridge.

V = fλ λ = \(\frac{v}{f}\) = \(\frac{3 \times 10^8}{600 \times 10^3}\) = 500m = 5 × 102m N.B 600KHz Frequency = 600 × 103Hz

Refraction, Interference and diffraction are all properties of waves

Nucleus of atom is made up of neutron and proton

When two mirror are inclined at an angle \theta, the number of images formed, n = \(\frac{360}{\theta}\) − 1 ∴ since \(\theta\) = 90o ∴ \(\frac{360}{\theta}\) − 1 = 4 − 1 = 3

For a refractive index () = \(\frac{\sin\frac{1}{2} (A + D)}{\sin\frac{1}{2}A}\) D = angle of minimum deviation A = refractive angle of the prism 1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin\frac{1}{2} \times 60}\) 1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin 30}\) Sin \(\frac{1}{2}\) (60 + D) = 1.5 * Sin 30 Sin \(\frac{1}{2}\) (60 + D) = 0.75 \(\frac{1}{2}\) (60 + D) = Sin-1 (0.75) but Sin-1 (0.75) = 49o \(\frac{1}{2}\) (60 + D) = 49 60 + D = 2 * 49 = 98o D = 98o - 60o D = 38o

Mechanical waves require material medium for their propagation only radio waves among the options docs require a material medium

It is a type of beam balance with great precision, and it can pick up the slightest mass placed on it.

This is a type of constructive interference. When two waves of equal amplitude are interfering constructively, the resulting amplitude is twice as large as the amplitude of an individual wave. = A + A = 2A

Dispersion occurs when different wavelengths of light are refracted by different amounts as they pass through the water droplets, causing the light to spread out into its constituent colors, thereby creating a rainbow.

Comparing the given equation with the wave equation: \(Asin(kx - \omega t)\), we have that A = 10 which is the amplitude of the wave.

\(\eta = \frac{1}{sin\theta}\) \(1.5 = \frac{1}{sin\theta}\) \(sin\theta = \frac{1}{1.5} = 0.667\) \(\theta = sin^{-1}(0.667) = 41.835^\circ\) \(\approxeq 41.8^\circ\)

For the 4th harmonic, \(L = 2\lambda\) \(4m = 2\lambda \implies \lambda = \frac{4m}{2}\) =\(2m\) Recall, \(v = f\lambda\) \(1 = f \times 2 \implies f = \frac{1}{2}\) = 0.50Hz

The number of images formed = \(\frac{360}{n}\) where the value gotten is odd else, = \(\frac{360}{n} -1\). Solving, we have that \(\frac{360}{20} = 18\) Hence, the no of images formed is 18 - 1 = 17.