(a) Explain the term critical angle. (b) List two factors which determine the deviation of a ray of light by a triangular glass prism. (c) The angle of refr...

(a) Critical angle. The critical angle is the angle of incidence in the optically denser medium for which the corresponding angle of refraction in the less dense medium is exactly \(90^\circ\). For an angle of incidence greater than this value, total internal reflection occurs.

(b) Two factors that determine the deviation by a triangular prism.

• The angle of incidence of the ray on the first face.
• The refracting (apex) angle of the prism, and the refractive index of the glass (which itself depends on the colour/wavelength of the light).

(c) Least deviation. The ray is least deviated when it passes symmetrically through the prism, so the two internal angles are equal to the given angle of refraction: \(r_1=r_2=r=29.0^\circ\). The refracting angle is then

\[A=r_1+r_2=2(29.0^\circ)=58.0^\circ.\]

At the first face, using \(n=\dfrac{\sin i}{\sin r}\) with \(n=1.5\):

\[\sin i = n\sin r = 1.5\times\sin 29.0^\circ = 1.5\times0.4848 = 0.7272,\]

\[i=\sin^{-1}(0.7272)=46.7^\circ.\]

The angle of minimum (least) deviation is

\[D_{min}=2i-A = 2(46.7^\circ)-58.0^\circ = 93.4^\circ-58.0^\circ \approx 35.4^\circ.\]

(d) From the ray diagram the apex angle \(A=60^\circ\), the angle of incidence on the left face is \(45^\circ\), and the ray is refracted through \(30^\circ\) inside the glass.

(i) Angles P, Q and R.

\(P\) is the angle of refraction at the first face, read directly from the diagram: \(P=r_1=30^\circ\).

Since \(r_1+r_2=A\), the angle of incidence at the second face is \(R=r_2=A-r_1=60^\circ-30^\circ=30^\circ\).

\(Q\) is the angle between the two normals (drawn at P and R) where they meet inside the prism. In the triangle formed by the internal ray PR and the two normals, \[P+R+Q=180^\circ,\] \[Q=180^\circ-30^\circ-30^\circ=120^\circ.\]

So \(P=30^\circ,\; Q=120^\circ,\; R=30^\circ.\)

(ii) Refractive index n. At the first face,

\[n=\frac{\sin 45^\circ}{\sin 30^\circ}=\frac{0.7071}{0.5000}=1.41.\]

(iii) Angle of emergence e. At the second face the angle of incidence inside is \(r_2=30^\circ\), so

\[\sin e = n\sin r_2 = 1.41\times\sin 30^\circ = 1.41\times0.5 = 0.707,\]

\[e=\sin^{-1}(0.707)=45^\circ.\]

(iv) Total deviation D.

\[D=(i_1+e)-A=(45^\circ+45^\circ)-60^\circ=30^\circ.\]

Because \(i_1=e=45^\circ\) and \(r_1=r_2=30^\circ\), the ray passes symmetrically, so this \(30^\circ\) is in fact the minimum deviation.