Question

10(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) Replacing the lens by another converging lens. Since the object moved 25 cm further away, it has its sharp image on the screen. Determine the focal length of the second lens.

Accepted Answer

(a) The critical angle is defined as an angle of incidence where the angle of refraction is 90°. This usually happens when light travels from a denser medium to a less dense medium.

(b) Antinodes are created in a stationary wave as a result of the superposition of the reflected waves. It is a result of the interference between two waves traveling in the opposite direction.

(c) Given: Angle of minimum deviation Dm=46.2°, required refractive index n=?
n = sin((Dm+A)/2) / sin(A/2)
For an equilateral triangle, A=60°
n = sin((46.2+60)/2) / sin(60/2) = sin(53.1°)/sin(30°) = 0.7997/0.5 = 1.5993 ≈ 1.6

(d) Possible reasons:
(i) The concave mirror and the screen may not be coaxially placed.
(ii) The illuminated object might be placed at a distance less than the focal length of the mirror (making the image virtual, formed behind the mirror, so it cannot appear on the screen).

(e)(i) Given: f=30cm, u=75cm, find v.
1/f = 1/u + 1/v
1/30 = 1/75 + 1/v
1/v = 1/30 - 1/75 = (5-2)/150 = 3/150
v = 150/3 = 50 cm

(ii) The object moved 25cm further away, so new u = 75+25 = 100cm, while v remains 50cm.
1/f = 1/u + 1/v = 1/100 + 1/50 = 1/100 + 2/100 = 3/100
f = 100/3 = 33.33 cm

WAEC Physics 2016 Theory — Question 10

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