FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Saturday, November 30, 2019

Apparatus used to produce interference fringes is shown in Fig. 6.1. The apparatus is not drawn to scale.


Question 28
(a) Apparatus used to produce interference fringes is shown in Fig. 6.1. The apparatus is
not drawn to scale.


Fig. 6.1 (not to scale)

Laser light is incident on two slits. The laser provides light of a single wavelength.
The light from the two slits produces a fringe pattern on the screen. A bright fringe is
produced at C and the next bright fringe is at B. A dark fringe is produced at P.

(i) Explain why one laser and two slits are used, instead of two lasers, to produce a
visible fringe pattern on the screen. [1]

(ii) State the phase difference between the waves that meet at
1. B [1]
2. P [1]

(iii) 1. State the principle of superposition. [2]
2. Use the principle of superposition to explain the dark fringe at P. [1]


(b) In Fig. 6.1 the distance from the two slits to the screen is 1.8 m. The distance CP is
2.3 mm and the distance between the slits is 0.25 mm.
Calculate the wavelength of the light provided by the laser. [3]





Reference: Past Exam Paper – June 2011 Paper 22 Q6





Solution:
(a)
(i) To produce coherent sources or constant phase difference.

(ii)
1. 360o / 2Ï€      allow n × (360o)          or n × (2Ï€)
{A bright fringe is seen due to constructive interference. So, the waves from the slits should be in phase. The phase difference is thus 360o / 2Ï€ or a multiple of it – that is, the 2 waves are completely in phase. A phase difference of 360o / 2Ï€ means no phase difference.

Note that the phase difference of 360° means that the waves are in phase. It produces the same effect as two waves having a phase difference of 0 (that is, they are also in phase).

2. 180o / Ï€        allow (n × 360o) – 180o           or (n × 2Ï€) – Ï€
{A dark fringe is seen due to destructive interference. So, the waves from the slits should be out of phase. The phase difference is thus 180o / Ï€ or (n × 360o) – 180o or (n × 2Ï€) – Ï€ – that is, the 2 waves are completely out of phase.}

(iii) 1. The principle of superposition states that when waves overlap / meet, the (resultant) displacement is the sum of displacements of each wave.

2. At P, there is a crest on a trough


(b)
{CP is the width between a dark fringe and a bright fringe. In the equation below, the fringe width is the distance between 2 successive bright fringe or 2 successive dark fringe. So, we need to consider twice the distance CP.}

Wavelength λ = ax / D
Wavelength λ = (2× 2.3×10-3) × (0.25×10-3) / 1.8
Wavelength λ = 639 nm

Friday, November 29, 2019

The diagram shows a non-uniform electric field near a positively charged and a negatively charged sphere.


Question 25
The diagram shows a non-uniform electric field near a positively charged and a negatively charged sphere.

Four electrons, A, B, C and D, are shown at different positions in the field.

On which electron is the direction of the force on the electron shown correctly?







Reference: Past Exam Paper – November 2011 Paper 12 Q31





Solution:
Answer: A.

The direction of electric field lines is the direction of the force on a positive charge. That is, the direction is from (away) a positive charge towards a negative charge. It represents the force acting on a positive charge (a positive charge would be attracted towards a negative charge).

An electron, which has a negative charge, will be attracted towards the positive sphere [B incorrect], NOT towards the negative charge [D incorrect] since like charges repel and unlike charges attract.


The resultant field between the two charges is non-uniform.

In a non-uniform electric field, the direction of the electric field on a charge at any specific point is given by tangent to the electric field line at that point in the electric field. [C incorrect since it is not a tangent] 

Choice C represents the direction of the force on the charge if only the positive charge was present.
Currently Viewing: Physics Reference | November 2019