• Physics 9702 Doubts | Help Page 66

Question 366: [Waves]
Monochromatic plane wave of speed c and wavelength λ is diffracted at small aperture.
Diagram illustrates successive wavefronts.

After what time will some portion of wavefront XY reach point P?
A 3λ / 2c                     B 2λ / c                        C 3λ / c                        D 4λ / c

Reference: Past Exam Paper – June 2012 Paper 12 Q29



Solution 366:
Answer: C.
The distance between the 2 successive lines (wavefronts) is λ. This is for both the straight and curved lines. So, wavefront XY is separated by a distance of 3λ from the wavefront that has reached point P.

Distance that needs to be travelled = 3λ
Speed of light = c

Speed = Distance / Time
Time = distance / speed = 3λ / c










Question 367: [Waves > Diffraction]

(a) State what is meant by diffraction of a wave.

(b) Plane wavefronts are incident on a slit, as shown in Fig.

Complete Fig. to show four wavefronts that have emerged from the slit.

(c) Monochromatic light is incident normally on diffraction grating having 650 lines per millimetre, as shown.

An image (the zero order) is observed for light that has angle of diffraction equal to zero.

For incident light of wavelength 590 nm, determine number of orders of diffracted light that can be observed on each side of the zero order.

(d) Images in Fig. are viewed, starting with the zero order and then with increasing order number.

State how appearance of the images changes as the order number increases.

Reference: Past Exam Paper – November 2010 Paper 21 Q5



Solution 367:

(a) When a wave passes through a slit / by an edge, the wave spreads out / changes direction.

(b) In the diagram, the wavelength is unchanged (separation lines is same and equal) and the wavefront is flat at the centre, curving into geometrical shadow.

(c)

For a diffraction grating,

d sinθ = nλ

{The angle for the largest order is 90^o.}

For angle of diffraction, θ = 90⁰,

{There are 650 in 1mm (= 0.001m = 10^-3m). Slit separation, d = separation between 2 successive lines = 10^-3 / 650 = 1 / (650x10³) m. Also, sin 90^o = 1.}

1 / (650x10³) = n (590x10^-9)

n = 2.6

{The number of order can only be an integer.}

Number of orders that can be observed is 2

(d) The intensity / brightness decreases (as the order increases).

Question 368: [Current of Electricity > Resistance]
Diagram shows a low-voltage circuit for heating water in a fish tank.

Heater has a resistance of 3.0 Ω. Voltage source has an e.m.f. of 12 V and internal resistance of 1.0 Ω.
At what rate does voltage source supply energy to the heater?
A 27 W                       B 36 W                        C 48 W                        D 64 W

Reference: Past Exam Paper – June 2006 Paper 1 Q34



Solution 368:

Answer: A.

Rate of energy supplied to heater = Power dissipated by heater

Power P dissipated by = I²R               where R is the resistance of the heater

Total resistance in circuit, R_T = 3.0 + 1.0 = 4.0 Ω

Current I in circuit = V / R_T = 12 / 4.0 = 3.0 A

Power dissipated by heater = I²R = (3.0)² (3.0) = 27 W  

Question 369: [Current of Electricity]
Four point charges, each of charge Q, are placed on edge of an insulating disc of radius r.
Frequency of rotation of the disc is f.

What is the equivalent electric current at the edge of disc?
A 4Qf              B 4Q / f                       C 8πrQf                       D 2Qf / πr

Reference: Past Exam Paper – November 2006 Paper 1 Q31



Solution 369:
Answer: A.
Total charge = It         where I is the current and t is the time

Total charge on disc = Q + Q + Q + Q = 4Q

Frequency = 1 / Period for 1 rotation
Period t for 1 rotation = 1 / f

Current I = Total charge / time = 4Q / (1/f) = 4Qf