(a) Explain (i) refraction of a wave; (ii) critical angle (b) State two conditions necessary for (i) total internal reflection of a wave to occur (ii) inter...

Assessment: WAEC SSCE - Physics - 2005 (Essay) Subject: Physics

Question 1 Report

(a) Explain (i) refraction of a wave;

(ii) critical angle

 

(b) State two conditions necessary for (i) total internal reflection of a wave to occur

(ii) interference wave patterns to be formed

 

(c) The distance between two successive crests of a water wave travelling at 3.6ms\(^{-1}\) is 0.45m, calculate the frequency of the wave

 

(d) A ray of light is incident at an angle of 30° at an air-glass interface.

(i) Draw a ray diagram to show the deviation of the ray in the glass.

(ii) Determine the angle of deviation. [Refractive index of glass = 1.50]

Answer Details

(a)

(i) Refraction of a wave is the change in direction of a wave when it passes obliquely from one medium to another in which its speed is different.

(ii) Critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is \(90^\circ\).

(b)(i) Conditions for total internal reflection

  1. The wave must travel from an optically denser medium to an optically less dense medium.
  2. The angle of incidence in the denser medium must be greater than the critical angle.

(b)(ii) Conditions for formation of an interference pattern

  1. There must be two or more coherent wave sources, that is, sources of the same frequency having a constant phase difference.
  2. The sources must be close together, with their separation of the order of the wavelength, so that the waves overlap.

(c)

The distance between successive crests is the wavelength:

\[\lambda=0.45\,\text{m},\qquad v=3.6\,\text{m s}^{-1}\]

Using \(v=f\lambda\),

\[f=\frac{v}{\lambda}=\frac{3.6}{0.45}=8.0\,\text{Hz}.\]

(d)(i) The ray bends towards the normal as it enters the glass, which is optically denser than air.

figure
Ray diagram for refraction from air into glass. The refracted ray bends towards the normal.

(d)(ii)

By Snell's law,

\[n=\frac{\sin i}{\sin r}\]

\[\sin r=\frac{\sin30^\circ}{1.50}=\frac{0.5}{1.50}=0.3333\]

\[r=\sin^{-1}(0.3333)=19.5^\circ\]

Therefore, the angle of deviation is

\[d=i-r=30.0^\circ-19.5^\circ=10.5^\circ.\]

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