• Physics 9702 Doubts | Help Page 137

Question 678: [Electric field]

An electric field exists in space between two charged metal plates.

Which of the following graphs shows variation of electric field strength E with distance d from X along the line XY?

Reference: Past Exam Paper – June 2003 Paper 1 Q35 & November 2006 Paper 1 Q30 & June 2012 Paper 12 Q31

Solution 678:

Answer: C.

Electric field strength = V / x

where V is the p.d. between the 2 plates and x is the separation of the plates.

Since the separation of the plates and the p.d. between the plates are constant, the electric field strength is also constant anywhere between the plates.

Question 679: [Waves > Double Slit experiment]

Laser is placed in front of a double slit, as shown in Fig.1. 

Laser emits light of frequency 670 THz. Interference fringes are observed on screen.

(a) Explain how interference fringes are formed

(b) Show that wavelength of the light is 450nm

(c) Separation of the maxima P and Q observed on screen is 12 mm. Distance between the double slit and the screen is 2.8 m.

Calculate separation of the two slits

(d) Laser is replaced by a laser emitting red light. State and explain the effect on the interference fringes seen on screen

Reference: Past Exam Paper – June 2014 Paper 22 Q7

Solution 679:

(a) The waves from the double slit are coherent / constant phase difference. The waves (from each slit) overlap / superpose / meet.

Maximum / bright fringes are observed where the path difference is nλ OR phase difference is n(360^o) / 2πn rad

OR Minimum / dark fringes are observed where the path difference is (n + ½)λ OR phase difference is (2n + 1)180^o / (2n + 1)π rad

(b)

v = fλ 

So, wavelength, λ = (3x10⁸) / (670x10¹²) = 448 (or 450) nm

(c)

{12mm represents the separation of 9 bright fringes (from the diagram). P and Q are bright fringes. Note that if we include the fringes at P and Q, there is 10 fringes – but we need to consider the separation here.}

Separation of bright fringes = 12 / 9 mm

Separation of 2 slits, a (= Dλ/w) = (28)(450x10^-9) / ({12/9}x10^-3) = 9.5x10^-4m

(d) Red light has a larger / higher / longer wavelength (must be comparison). So, the fringes are further apart / larger separation

Question 680: [Measurement > CRO]

A microphone detects musical note of frequency f. The microphone is connected to cathode-ray oscilloscope (c.r.o.). Signal from the microphone is observed on the c.r.o. as illustrated in Fig.1.

Time-base setting of the c.r.o. is 0.50 ms cm^–1. Y-plate setting is 2.5 mV cm^–1.

(a) Use Fig.1 to determine

(i) amplitude of the signal,

(ii) frequency f,

(iii) actual uncertainty in f caused by reading the scale on the c.r.o.

(b) State f with its actual uncertainty.

Reference: Past Exam Paper – November 2014 Paper 23 Q2

Solution 680:

(a)

(i)

{On the scale, the amplitude is distance of 2.2cm from the zero line.}

Amplitude scale reading is 2.2 (cm)

{The Y-plate setting is 2.5 mV cm^–1. That is, 1cm corresponds to 2.5 mV.}

Amplitude = 2.2 × 2.5 = 5.5 mV

(ii)

Time period scale reading = 3.8 (cm)

{Time-base setting of the c.r.o. is 0.50 ms cm^–1.}

Time period = 3.8 × (0.5 × 10^–3) = 0.0019 (s)

{Frequency = 1 / period}

Frequency f = 1 / 0.0019 = 530 (526) Hz

(iii)

{We need to find the actual uncertainty in the frequency f.

The uncertainty in the reading of the scale corresponds to 1 small square. Note that 5 small squares is 1cm, as indicated on the diagram. So, 1 small square is 0.2cm.}

Time period scale reading = 3.8 (cm). This corresponds to a period of 0.0019 (s).

EITEHR

If we measure from the zero position to 1 period, the value read is 3.8cm on the scale and the uncertainty is 0.2cm. This corresponds to a percentage uncertainty of (0.2 / 3.8) x 100% = 5.3%

OR

If we measure from the zero position to 2 periods (measuring more periods reduces the uncertainty), the value read is 7.6cm on the scale and the uncertainty is 0.2cm. This corresponds to a percentage uncertainty of (0.2 / 7.6) x 100% = 2.6%}

EITHER Uncertainty in reading = ± 0.2 in 3.8 (cm) OR 5.3%

OR Uncertainty in reading = ± 0.2 in 7.6 (cm) OR 2.6%

{Frequency f = 1 / period T. So, there will be the same percentage uncertainty in the frequency, as the percentage uncertainty in the period. The frequency f was calculated to be 530 (526) Hz.

EITHER Uncertainty in f = 5.3% of 526       OR 2.6% of 526}

EITHER Actual uncertainty = 5.3% of 526 = 27.7 or 28 Hz

OR Actual uncertainty = 2.6% of 526 = 13 or 14

(b) State f with its actual uncertainty.

{The uncertainty should be expressed to one significant figure with the frequency quoted to the same power-of-ten as the uncertainty.}

Frequency = 530 ± 30 Hz or 530 ± 10 Hz

Question 681: [Electric field]

Diagram shows an electron, with charge e, mass m, and velocity v, entering a uniform electric field of strength E.

Direction of the field and the electron’s motion are both horizontal and to the right.

Which expression gives the distance x through which electron travels before it stops momentarily?

A x = mv / E               B x = mv / Ee              C x = mv² / 2E            D x = mv² / 2Ee

Reference: Past Exam Paper – June 2009 Paper 1 Q29

Solution 681:

Answer: D.

Electric force on the electron = Ee

This force is the resultant force on the electron and is equal to ma, where a is the acceleration.  The resultant force on the electron causes it to decelerate until it comes to rest.

So,

Magnitude of acceleration a = Ee / m

(The acceleration of the electron should be taken as negative as it opposes motion – that is, it is a deceleration.)

Equation for uniformly accelerated motion: v² = u² + 2as

Initial velocity of electron (corresponding to u in the equation above) is v (as given in the question). Final velocity of the electron = 0.

0 = v² + 2(-a)x

Re-arranging and substituting the value of a in the equation gives

v² = 2 (Ee/m) x

Thus, x = mv² / 2Ee