a) (i) Illustrate, using a ray… | Physics

Question 1 Report

a) (i) Illustrate, using a ray diagram, how an image can be formed by a convex mirror.

(ii) State one advantage rid one disadvantage of using a convex mirror as a iving mirror.

(iii) Explain the action of a compound microscope.

(b) Illustrate using labelled diagrams only, sonometer wire of length I, vibrating at its fundamental (ii) first overtone (iii) second overtone

(c) A tuning fork vibrating at a frequency of 512 Hz is held over the top of a jar filled with water and fitted with a tap at the buttom. If the jar is 60 cm tall and the speed of sound is 350 ms\(^{-1}\), determine the possible resonance position(s).

Answer

(a)(i) Formation of an image by a convex mirror

Ray diagram showing the formation of a virtual, erect and diminished image by a convex mirror.

The reflected rays diverge, but their backward extensions meet behind the mirror. Hence the image is virtual, erect and diminished, and is formed between the pole, P, and the principal focus, F.

(a)(ii) Convex mirror as a driving mirror

Advantage: It provides a wide field of view and gives an erect image of vehicles behind.
Disadvantage: The image is diminished, so objects appear farther away than they actually are and distance may be misjudged.

(a)(iii) Action of a compound microscope

A compound microscope consists of two converging lenses of short focal lengths: the objective lens and the eyepiece lens. The object is placed just beyond the focal point of the objective lens. The objective forms a real, inverted and magnified intermediate image within the focal length of the eyepiece. The eyepiece acts as a simple magnifying glass and produces a final image which is virtual, highly magnified and inverted relative to the original object.

(b) Modes of vibration of a sonometer wire

Labelled diagrams of a sonometer wire vibrating in its fundamental mode, first overtone and second overtone. N denotes a node and A denotes an antinode.

(c) Resonance positions

The air column above the water is closed at the water surface and open at the top. Thus, resonance occurs at air-column lengths:

\[L=\frac{\lambda}{4},\quad \frac{3\lambda}{4},\quad \frac{5\lambda}{4},\ldots\]

\[\lambda=\frac{v}{f}=\frac{350}{512}=0.6836\ \text{m}\approx0.684\ \text{m}\]

First resonance position:

\[L_1=\frac{\lambda}{4}=\frac{0.6836}{4}=0.1709\ \text{m}\approx17.1\ \text{cm}\]

Second resonance position:

\[L_2=\frac{3\lambda}{4}=\frac{3(0.6836)}{4}=0.5127\ \text{m}\approx51.3\ \text{cm}\]

Third resonance position:

\[L_3=\frac{5\lambda}{4}=\frac{5(0.6836)}{4}=0.8545\ \text{m}\approx85.5\ \text{cm}\]

Since the jar is only \(60\ \text{cm}\) high, the third resonance is not possible. Therefore, the possible resonance air-column lengths are 17.1 cm and 51.3 cm, measured downwards from the open top of the jar.

Read lesson note on Reflection Of Light At Plane And Curved Surfaces (JAMB)

Read lesson note on Light Waves (WAEC)

Answer Details