• Physics 9702 Doubts | Help Page 159

Question 787: [Waves > Diffraction]

Interference patterns from a diffraction grating and a double slit are compared.

Using diffraction grating, yellow light of the first order is seen at 30° to the normal to the grating.

The same light produces interference fringes on screen 1.0 m from the double slit. The slit separation is 500 times greater than the line spacing of the grating.

What is the fringe separation on the screen?

A 2.5 × 10^–7 m

B 1.0 × 10^–5 m

C 1.0 × 10^–3 m

D 1.0 × 10^–1 m

Reference: Past Exam Paper – November 2006 Paper 1 Q27

Solution 787:

Answer: C.

For double slits: Separation of slits, a = Dλ / w

where D is the distance of the slit form the screen, w is fringe separation and λ is the wavelength

For diffraction grating: d sinθ = n λ

Since the same light is used, the wavelength λ is the same in both cases.

For the diffraction grating, the 1^st order (n = 1) is seen at 30° to the normal (θ = 30°)

d sin30° = 1 λ

Wavelength λ = 0.5d

For double slits,

Separation of slits, a = Dλ / w

Fringe separation, w = Dλ / a

In the question, it is stated that the slit separation (a) for the double slit is 500 times greater than the line spacing of the grating (d).

a = 500d

Wavelength λ = 0.5d

Fringe separation, w = Dλ / a = D (0.5d) / 500d = 0.5 (1.0) / 500 = 1.0 × 10^–3 m

Question 788: [Matter > Young modulus]

Behaviour of a wire under tensile stress may be described in terms of the Young modulus E of the material of the wire and of the force per unit extension k of the wire.

For wire of length L and cross-sectional area A, what is the relation between E and k?

A E = A / kL               B E = kA / L               C E = kL / A               D E = L/ kA

Reference: Past Exam Paper – June 2011 Paper 12 Q23

Solution 788:

Answer: C.

Young modulus E = stress / strain

Stress = Force / Area = F / A

Strain = extension / original length = e / L

Young modulus E = (F/A) / (e/L) = FL / Ae

But Hooke’s law:  F = ke giving k = F/e.

Young modulus E = kL / A

Question 789: [Current of Electricity]

Diagram shows the circuit for a signal to display a green or a red light. It is controlled by the switch S.

The signal is some way from S to which it is connected by cable with green, red and black wires. At the signal, green and red wires are connected to the corresponding lamp and the black wire is connected to a terminal x to provide a common return. The arrangement is shown correctly connected and with the switch set to illuminate the red lamp.

During maintenance, wires at the signal are disconnected and, when reconnected, the black wire is connected in error to the green lamp (terminal g) instead of terminal x. Red wire is connected correctly to its lamp and connections at S remain as in the diagram.

When the system is tested with switch connection to the red wire, what does the signal show?

A the green lamp illuminated normally

B the red lamp illuminated normally

C the red and green lamps both illuminated normally

D the red and green lamps both illuminated dimly

Reference: Past Exam Paper – June 2012 Paper 12 Q37

Solution 789:

Answer: D.

The lamps are lit dimly since they are connected in series.

The same power supply is now providing current to both lights at the same time.

Ohm’s law: V = IR

The total resistance in the circuit is higher and thus the current flowing in the circuit is less.

Each of the lights receive a voltage lower than 12V such that the total voltage across the two of them is 12V. Previously, all the 12V was being provided to a single lamp.

Question 790: [Current of Electricity]

A cell of e.m.f. E and internal resistance r is connected in series with switch S and an external resistor of resistance R.

The p.d. between P and Q is V.

When S is closed,

A V decreases because there is a p.d. across R.

B V decreases because there is a p.d. across r.

C V remains the same because the decrease of p.d. across r is balanced by the increase of p.d. across R.

D V remains the same because the sum of the p.d.s across r and R is still equal to E.

Reference: Past Exam Paper – November 2011 Paper 11 Q36 & Paper 13 Q38

Solution 790:

Answer: B.

When the switch S is open, no current flows in the circuit. So, the p.d. between P and Q is equal to the e.m.f. E of the cell.

The cell has an internal resistance, so as the switch is closed (the circuit is now complete for a current to flow) there is some loss volt (V_L = Ir) across the internal resistance r of the cell. That is, the p.d. between P and Q is now less than the e.m.f. of the cell (V = E – V_L = E – Ir).

Question 791: [Vectors]

Diagram shows a resultant force and its horizontal and vertical components.

The horizontal component is 20.0 N and θ = 30°. What is the vertical component?

A 8.7 N                       B 10.0 N                     C 11.5 N                     D 17.3 N

Reference: Past Exam Paper – June 2009 Paper 1 Q2

Solution 791:

Answer: C.

Let the resultant vector be R.

Horizontal component = R cosθ

Vertical component = R sinθ

Horizontal component = R cosθ = 20.0 N

Resultant vector R = 20 / cosθ

Vertical component = R sinθ = (20/cosθ) sinθ = 20 tanθ = 20 tan30° = 11.5 N

Question 792: [Waves > Diffraction grating]

Monochromatic light is incident on diffraction grating and a diffraction pattern is observed.

Which line of table gives the effect of replacing the grating with one that has more lines per metre?

Reference: Past Exam Paper – June 2014 Paper 13 Q30

Solution 792:

Answer: B.

Diffraction grating: dsinθ = nλ

Replacing the grating with one that has more lines per metre causes the slit separation, d to decrease.

So, the number of orders of diffraction visible, n (= dsinθ / λ) decreases [since d is decreasing] while the angle between first and second orders of diffraction, θ (= sin^-1(λ/d) and can only be between 0 and 90) increases [between 0 and 90, the greater the argument of sin^-1, the closer the angle is to 90 {as d decreases, the argument increases}].