(a) Define critical angle.
(b) How are anti-nodes created in a stationary wave?
(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 is defined as the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90°.
(b) Antinodes are created in a stationary wave when an incident and its reflected waves are superposed and 180° out of phase.
(c) n = \(\frac{Sin\frac{(A + Dm)}{2}}{\sin (\frac{A}{2})}\)
= \(\frac{Sin\frac{(60 + 46.2)}{2}}{\sin (\frac{60}{2})}\)
= \(\frac{sin 53.1}{sin 30}\)
= 1.6
(d) No image is obtained on the screen because
- Object, screen and mirror are not coaxial;
- Object is placed at focal point of mirror;
- Object is placed between the focus and the pole;
- Screen is located at a distance less than or greater than actual image distance.
(e) (i) \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{30} - \frac{1}{7}\)
= \(\frac{5 - 2}{150} = \frac{3}{150}\)
U = 75 + 25 = 100cm
\(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\)
\(\frac{1}{100} + \frac{1}{50} = \frac{1}{50} = \frac{1}{50}\)
f = 50cm
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