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WAEC Physics 2020 Theory — Question 10

Question 10 of 12 from the West African Examinations Council (WAEC) Physics 2020 Theory paper, with the correct answer and a full explanation.

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10(a)(i) Define critical angle as used in optics. (ii) State two conditions necessary for total internal reflection to occur. (iii) List three practical applications of total internal reflection. (b) State two effects of refraction. (c)(i) Define progressive waves. (ii) A plane progressive wave is represented by the equation y=0.5sin(1000(pi)t - 100(pi)x/17), where y is in millimeters, t in seconds, and x in meters. Calculate the: (alpha) frequency of the wave; (beta) wavelength of the wave; (gamma) speed of the wave.

Model answer

(a)(i) Critical angle is defined as the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is 90 deg. i=c=critical angle when r=refracted angle=90 deg. (ii) Conditions necessary for total internal reflection to occur: light rays must travel from a denser medium to a less dense medium e.g. glass to air or glass to water or water to air; the angle of incidence in the denser medium must be greater than the critical angle (i>c). (iii) Practical applications of total internal reflection (TIR): Optical fibres used in the telecommunication field; Prism periscopes for viewing overhead objects; Prism Binoculars for viewing distance objects; Fish eye view; Transmission of radio signals. (b) Effects of refraction: Change in the direction of a wave; Change in the speed of a wave; Object appears raised when viewed above the water. (c)(i) A progressive wave is defined as a disturbance which travels through a medium that enables energy to be transferred from its source to another (without the particles of the medium being transferred). (ii) Given the wave equation y=0.5sin(1000(pi)t-100(pi)x/17): (alpha) To find the frequency of the wave, we recall the general equation of a progressive wave along the positive x-axis, i.e. y=Asin(omega*t - 2(pi)x/lambda) = Asin(2(pi)ft - 2(pi)x/lambda). Comparing this with the given equation, we have 2(pi)f=1000(pi), so f=1000(pi)/2(pi)=500 Hz. (beta) To find the wavelength of the wave, again we compare the two wave equations, i.e. 2(pi)x/lambda = 100(pi)x/17. So 2/lambda=100/17. lambda x 100 = 17x2. lambda=34/100=0.34 m. (gamma) To calculate the speed of the wave, using v=f*lambda: v=500x0.34, v is approximately 170 m/s.

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