# 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 nÎ»
OR phase difference is n(360

^{o}) / 2Ï€n rad
OR Minimum / dark fringes are observed
where the path difference is (n + ½)Î» OR phase difference is (2n + 1)180

^{o}/ (2n + 1)Ï€ rad**(b)**

v = fÎ»

So, wavelength, Î» = (3x10

^{8}) / (670x10^{12}) = 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 (= DÎ»/w) = (28)(450x10

^{-9}) / ({12/9}x10^{-3}) = 9.5x10^{-4}m**(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 = mv

^{2}/ 2E D x = mv^{2}/ 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: v

^{2}= u^{2}+ 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 = v

^{2}+ 2(-a)x
Re-arranging and substituting the
value of a in the equation gives

v

^{2}= 2 (Ee/m) x
Thus, x = mv

^{2}/ 2Ee
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ReplyDeleteSee question 683 at

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Here is the first question from paper 1:

ReplyDelete12/M/J/10 Q.6

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21/O/N/14 Q.3(d), Q.4(a)(iii), Q.8(b)(i)

ReplyDeleteFor question 8, see solution 726 at

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22/O/N/14 Q.1(c)(i)

ReplyDeleteSee solution 688 at

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May june 2003 p1 q 22

ReplyDeleteSee solution 849 at

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