(a) Define critical angle.
(c) The angle of minimum deviation of an equilateral triangular glass prism is 46.2°. Calculate the refractive index of the glass.
(d) An illuminated object is placed in front of a concave mirror and the position of a screen is adjusted in front of the mirror but no image is obtained on the screen. Give two possible reasons for this observation.
(e) An illuminated object is placed at a distance of 75 cm from a converging lens of focal length 30 cm.
(i) Determine the image distance.
(ii) If the lens is replaced by another converging lens, the object has to be moved 25 cm further away to have its sharp image on the screen. Determine the focal length of the second lens.
(a) Critical angle: the angle of incidence, in the denser medium, for which the angle of refraction in the less dense medium is \(90^{\circ}\). For angles greater than this, total internal reflection occurs.
(b) Formation of antinodes: antinodes are points of maximum displacement in a stationary wave. They are formed where the incident wave and the reflected wave meet in phase, so their displacements always add up (constructive superposition), giving maximum vibration.
(c) Refractive index of the prism. For a prism of refracting angle \(A=60^{\circ}\) (equilateral) at minimum deviation \(D=46.2^{\circ}\):
\[ n = \frac{\sin\!\left(\frac{A+D}{2}\right)}{\sin\!\left(\frac{A}{2}\right)} = \frac{\sin\!\left(\frac{60+46.2}{2}\right)}{\sin 30^{\circ}} = \frac{\sin 53.1^{\circ}}{0.5} = \frac{0.7997}{0.5} = 1.60 \]
(d) Concave mirror gives no image on the screen - two possible reasons:
- The object is placed at (or inside) the focal point, so the reflected rays are parallel or diverge and form a virtual image that cannot be caught on a screen.
- The screen is not placed at the correct image distance (it is nearer or farther than the position where the rays actually converge).
(e) Converging lens.
(i) \(u=75\,\text{cm}\), \(f=30\,\text{cm}\). Using \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\):
\[ \frac{1}{v}=\frac{1}{30}-\frac{1}{75}=\frac{5-2}{150}=\frac{3}{150}=\frac{1}{50}\Rightarrow v=50\,\text{cm} \]
(ii) The object is moved \(25\,\text{cm}\) farther, so \(u_2 = 75+25 = 100\,\text{cm}\), and the sharp image is formed on the same screen, i.e. \(v_2 = 50\,\text{cm}\).
\[ \frac{1}{f_2}=\frac{1}{u_2}+\frac{1}{v_2}=\frac{1}{100}+\frac{1}{50}=\frac{1+2}{100}=\frac{3}{100} \]\[ f_2 = \frac{100}{3} = 33.3\,\text{cm} \]
