Define diffraction.

Question

Define diffraction.

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Here are the solutions to all the questions, with explicit steps, explanations, and examples where possible.

1) Define diffraction.

Diffraction is the phenomenon where waves spread out as they pass through an aperture (opening) or around an obstacle. This spreading is most noticeable when the wavelength of the wave is comparable to the size of the aperture or obstacle.

2) A narrow beam of monochromatic light is incident normally on a diffraction grating. The third order diffraction beam makes an angle of $45^\circ$ with the undeviated beam. What is the maximum order beam of diffracted beam produced by the grating?

The diffraction grating equation is given by:

d \sin \theta = m \lambda

where $d$ is the grating spacing, $\theta$ is the diffraction angle, $m$ is the order of the maximum, and $\lambda$ is the wavelength of the light.

Step 1: Use the given information for the third order maximum to find the ratio $\frac{d}{\lambda}$ .
Given $m = 3$ and $\theta = 45^\circ$ .

d \sin 45^\circ = 3 \lambda

d \left(\frac{\sqrt{2}}{2}\right) = 3 \lambda

\frac{d}{\lambda} = \frac{3}{\sin 45^\circ} = \frac{3}{\frac{\sqrt{2}}{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}

\frac{d}{\lambda} \approx 3 \times 1.414 = 4.242

Step 2: Determine the maximum possible order ( $m_{max}$ ).
The maximum possible angle for diffraction is $\theta = 90^\circ$ , for which $\sin \theta = 1$ .
So, for the maximum order $m_{max}$ :

d \sin 90^\circ = m_{max} \lambda

d (1) = m_{max} \lambda

m_{max} = \frac{d}{\lambda}

Step 3: Substitute the value of $\frac{d}{\lambda}$ found in Step 1.

m_{max} = 3\sqrt{2} \approx 4.242

Since the order $m$ must be an integer, the maximum integer order of diffraction produced by the grating is the largest integer less than or equal to $m_{max}$ .

m_{max} = \lfloor 4.242 \rfloor = 4

The maximum order beam of diffracted beam produced by the grating is $\boxed{\text{4th order}}$ .

3) A diffraction grating has 400 lines per mm and is illuminated normally by monochromatic light of wavelength 600 nm. Calculate:

Given:

Number of lines per mm = 400 lines/mm
Wavelength $\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}$

a) The grating spacing

Step 1: Convert the number of lines per mm to lines per meter.

400 \, \text{lines/mm} = 400 \, \text{lines} / (1 \times 10^{-3} \, \text{m}) = 400 \times 10^3 \, \text{lines/m}

Step 2: Calculate the grating spacing $d$ .
The grating spacing $d$ is the reciprocal of the number of lines per unit length.

d = \frac{1}{\text{Number of lines per meter}}

d = \frac{1}{400 \times 10^3 \, \text{lines/m}} = \frac{1}{4 \times 10^5} \, \text{m}

d = 0.25 \times 10^{-5} \, \text{m} = 2.5 \times 10^{-6} \, \text{m}

The grating spacing is $\boxed{2.5 \times 10^{-6} \, \text{m}}$ .

b) The angle to the normal at which the first principal order is seen

Step 1: Use the diffraction grating equation $d \sin \theta = m \lambda$ .
For the first principal order, $m=1$ .

(2.5 \times 10^{-6} \, \text{m}) \sin \theta = 1 \times (600 \times 10^{-9} \, \text{m})

Step 2: Solve for $\sin \theta$ .

\sin \theta = \frac{600 \times 10^{-9}}{2.5 \times 10^{-6}} = \frac{600}{2500} = \frac{6}{25} = 0.24

Step 3: Calculate $\theta$ .

\theta = \arcsin(0.24) \approx 13.89^\circ

The angle to the normal at which the first principal order is seen is $\boxed{13.89^\circ}$ .

c) The number of diffraction maxima obtained

Step 1: Determine the maximum possible order $m_{max}$ .
The maximum angle for diffraction is $\theta = 90^\circ$ , so $\sin \theta = 1$ .

d \sin 90^\circ = m_{max} \lambda

m_{max} = \frac{d}{\lambda}

Step 2: Substitute the values of $d$ and $\lambda$ .

m_{max} = \frac{2.5 \times 10^{-6} \, \text{m}}{600 \times 10^{-9} \, \text{m}} = \frac{2.5 \times 10^{-6}}{0.6 \times 10^{-6}} = \frac{2.5}{0.6} \approx 4.167

Step 3: Find the largest integer order.
Since $m$ must be an integer, the maximum order observed on one side is $\lfloor 4.167 \rfloor = 4$ .
The diffraction maxima occur at $m = 0, \pm 1, \pm 2, \pm 3, \pm 4$ .
The total number of diffraction maxima (bright spots) obtained is $2m_{max} + 1$ .
Total maxima = $2(4) + 1 = 9$ .
If the question asks for the highest order, it's 4. If it asks for the total number of bright spots, it's 9. Given the phrasing "number of diffraction maxima obtained", it usually refers to the highest order.

The number of diffraction maxima obtained (highest order) is $\boxed{4}$ .

d) The slit width.

This question asks for "the slit width". In the context of a diffraction grating, the term "slit width" (often denoted by $a$ ) refers to the width of each individual opening in the grating, which is distinct from the grating spacing $d$ (the distance between the centers of adjacent slits). The problem provides information about the grating spacing $d$ and the wavelength $\lambda$ , but no information about the intensity distribution or the angles of diffraction minima that would allow us to determine the individual slit width $a$ .

Therefore, the slit width cannot be determined from the information provided.

4) a) Defined temperature coefficient of resistance

The temperature coefficient of resistance ( $\alpha$ ) is a material property that quantifies how much the electrical resistance of a substance changes per degree of temperature change. It is defined as the fractional change in resistance per unit change in temperature, usually referenced to a specific temperature (e.g., $0^\circ \text{C}$ or $20^\circ \text{C}$ ). A positive coefficient means resistance increases with temperature, while a negative coefficient means it decreases.

b) Describe an experiment to measure temperature coefficient of resistance. Your description should include a diagram, a procedure, some precautions, observations and a conclusion.

Experiment to Measure Temperature Coefficient of Resistance

Aim: To determine the temperature coefficient of resistance of a given wire (e.g., copper or nichrome).

Diagram:

       [Ammeter] ---- [Wire Coil (R_T)] ----
       |                                   |
       |                                   |
       [Power Supply]                      [Voltmeter]
       |                                   |
       |                                   |
       -------------------------------------
       (Wire coil immersed in a water bath with thermometer)

Note: A more precise setup would use a Wheatstone bridge for resistance measurement, but a simple voltmeter-ammeter method is also acceptable for demonstration.

Apparatus:

Coil of resistance wire (e.g., copper or nichrome)
Power supply (DC)
Ammeter
Voltmeter
Thermometer
Beaker or container for water bath
Heating plate or Bunsen burner
Stirrer
Connecting wires

Procedure:

Setup: Connect the resistance wire coil in series with an ammeter and a power supply. Connect a voltmeter in parallel across the resistance wire coil. Place the coil and a thermometer in a beaker filled with water, forming a water bath.
Initial Measurement: Record the initial temperature of the water bath ( $T_0$ ) using the thermometer. Turn on the power supply and adjust it to allow a small, steady current to flow through the coil. Record the current ( $I_0$ ) from the ammeter and the voltage ( $V_0$ ) from the voltmeter.
Calculate Initial Resistance: Calculate the resistance of the wire at $T_0$ using Ohm's Law: $R_0 = V_0 / I_0$ .
Heating and Measurement: Gradually heat the water bath using the heating plate, stirring continuously to ensure uniform temperature distribution. At regular temperature intervals (e.g., every $5^\circ \text{C}$ or $10^\circ \text{C}$ ), turn off the heater, allow the temperature to stabilize, and then record the new temperature ( $T$ ), current ( $I$ ), and voltage ( $V$ ).
Calculate Resistance at T: For each temperature, calculate the resistance $R_T = V / I$ .
Repeat: Continue heating and taking measurements until a suitable range of temperatures (e.g., up to $80^\circ \text{C}$ or $90^\circ \text{C}$ ) has been covered.
Plot Graph: Plot a graph of resistance ( $R_T$ ) against temperature ( $T$ ).

Observations:

As the temperature of the wire increases, its resistance will generally increase (for most metals like copper).
The graph of $R_T$ versus $T$ should be approximately a straight line.

Calculations and Conclusion:

Determine Slope: From the graph of $R_T$ vs $T$ , determine the slope ( $m = \frac{\Delta R}{\Delta T}$ ).
Calculate Temperature Coefficient: The temperature coefficient of resistance ( $\alpha$ ) can be calculated using the formula: $\alpha = \frac{1}{R_0} \left(\frac{\Delta R}{\Delta T}\right)$ where $R_0$ is the resistance at the reference temperature (e.g., the initial temperature $T_0$ or the intercept of the graph with the resistance axis at $0^\circ \text{C}$ ).
Conclusion: State the calculated value of the temperature coefficient of resistance for the given wire and compare it with known values for the material.

Precautions:

Low Current: Use a low current to prevent self-heating of the wire, which could lead to inaccurate temperature readings.
Stirring: Stir the water bath continuously to ensure uniform temperature distribution throughout the water and the coil.
Insulation: Ensure the resistance wire is properly insulated to prevent short circuits in the water bath.
Accurate Readings: Take careful readings of temperature, voltage, and current to minimize experimental errors.
Reference Temperature: Ensure the reference temperature $T_0$ is accurately measured.

5) In the circuit in the figure below a dry cell is connected to a rheostat of resistance R and an ammeter to measure current I in the circuit. When $R=10.0\Omega$ , $I=0.5A$ and if $R=22\Omega$ , $I=0.025A$ . Calculate the internal resistance of the cell?

The electromotive force (EMF) of a real cell is given by the formula:

E = I(R + r)

where $E$ is the EMF, $I$ is the current, $R$ is the external resistance, and $r$ is the internal resistance.

We are given two sets of data:

Case 1: $R_1 = 10.0 \, \Omega$ , $I_1 = 0.5 \, A$
Case 2: $R_2 = 22 \, \Omega$ , $I_2 = 0.025 \, A$

Assumption: Using the given value $I_2 = 0.025 \, A$ leads to a negative internal resistance, which is physically impossible. It is highly probable that $I_2$ was intended to be $0.25 \, A$ . We will proceed with this assumption to find a physically meaningful answer.

Step 1: Write down the equations for each case using the formula $E = I(R + r)$ .
For Case 1:

E = I_1(R_1 + r)

E = 0.5 \, A (10.0 \, \Omega + r)

E = 5.0 + 0.5r \quad (*)

For Case 2 (using the assumed value $I_2 = 0.25 \, A$ ):

E = I_2(R_2 + r)

E = 0.25 \, A (22 \, \Omega + r)

E = 5.5 + 0.25r \quad (**)

Step 2: Equate the two expressions for $E$ to solve for $r$ .
Since the EMF $E$ of the cell is constant:

5.0 + 0.5r = 5.5 + 0.25r

Step 3: Rearrange the equation to solve for $r$ .

0.5r - 0.25r = 5.5 - 5.0

0.25r = 0.5

r = \frac{0.5}{0.25}

r = 2 \, \Omega

Step 4: (Optional) Calculate the EMF $E$ using the value of $r$ to verify consistency.
Using equation $(*)$ :

E = 5.0 + 0.5(2) = 5.0 + 1.0 = 6.0 \, V

Using equation $(**)$ :

E = 5.5 + 0.25(2) = 5.5 + 0.5 = 6.0 \, V

The EMF is consistent.

The internal resistance of the cell is $\boxed{\text{2.0 } \Omega}$ .

6) A voltmeter, designed to measure voltages from zero to 10V, draws a current of 0.2A at full-scale deflection. To convert this to measure voltage up to 100V, it would be necessary to connect a resistor to this metal. What is the value of this resistor and how should it be connected?

Step 1: Calculate the internal resistance of the voltmeter ( $R_v$ ).
At full-scale deflection, the voltmeter measures $V_{max} = 10 \, V$ and draws a current $I_{fsd} = 0.2 \, A$ .
Using Ohm's Law, the internal resistance of the voltmeter is:

R_v = \frac{V_{max}}{I_{fsd}}

R_v = \frac{10 \, V}{0.2 \, A} = 50 \, \Omega

Step 2: Determine the total resistance required for the new range.
To measure a new maximum voltage $V'_{max} = 100 \, V$ with the same full-scale deflection current $I_{fsd} = 0.2 \, A$ , the total resistance of the voltmeter circuit ( $R_{total}$ ) must be:

R_{total} = \frac{V'_{max}}{I_{fsd}}

R_{total} = \frac{100 \, V}{0.2 \, A} = 500 \, \Omega

Step 3: Calculate the value of the series resistor ( $R_s$ ).
To increase the range of a voltmeter, a resistor must be connected in series with its internal resistance.

R_{total} = R_v + R_s

500 \, \Omega = 50 \, \Omega + R_s

R_s = 500 \, \Omega - 50 \, \Omega = 450 \, \Omega

Step 4: State how the resistor should be connected.
The resistor must be connected in series with the voltmeter.

The value of the resistor is $\boxed{450 \, \Omega}$ and it should be connected $\boxed{\text{in series}}$ with the voltmeter.

7) A battery of emf 24V of negligible internal resistance is connected as shown in the figure below. The reading of the ammeter is 2.0A. what is the value of the resistor R in the circuit

Given:

EMF $E = 24 \, V$
Total current $I_{total} = 2.0 \, A$
Resistors: $6.0 \, \Omega$ , $3.0 \, \Omega$ , and $R$ .

Step 1: Calculate the equivalent resistance of the parallel combination.
The $6.0 \, \Omega$ and $3.0 \, \Omega$ resistors are connected in parallel. Let their equivalent resistance be $R_p$ .

\frac{1}{R_p} = \frac{1}{6.0 , \Omega} + \frac{1}{3.0 , \Omega}

\frac{1}{R_p} = \frac{1}{6.0 , \Omega} + \frac{2}{6.0 , \Omega} = \frac{3}{6.0 , \Omega} = \frac{1}{2.0 , \Omega}

R_p = 2.0 , \Omega


Step 2: Calculate the total equivalent resistance of the circuit.  
The resistor $R$ is in series with the parallel combination $R_p$.

R_{total} = R + R_p

R_{total} = R + 2.0 , \Omega

Step 3: Use Ohm's Law for the entire circuit. Since the internal resistance of the battery is negligible, the total voltage across the circuit is the EMF.

E = I_{total} \times R_{total}

24 , V = 2.0 , A \times (R + 2.0 , \Omega)


Step 4: Solve for $R$.

\frac{24 , V}{2.0 , A} = R + 2.0 , \Omega

12 , \Omega = R + 2.0 , \Omega

R = 12 , \Omega - 2.0 , \Omega

R = 10 , \Omega


The value of the resistor R in the circuit is $\boxed{10 \, \Omega}$.

---

**8)** **The figure shows how three resistors, three batteries with their resistances and ammeters are connected in a network. Calculate the:**

**i) reading of the meters A, B and C**

**ii) Potential difference between M and N**

**iii) power dissipated as heat in the $5\Omega$ resistor**

This is a complex circuit that requires Kirchhoff's laws. Let's define the currents in each branch.  
Let $I_A$ be the current through ammeter A (downwards through the $4\Omega$ resistor).  
Let $I_B$ be the current through ammeter B (clockwise in the left loop).  
Let $I_C$ be the current through ammeter C (counter-clockwise in the right loop).

We can assign loop currents. Let $I_1$ be the current in the left loop (clockwise) and $I_2$ be the current in the right loop (clockwise).  
Then:  
$I_B = I_1$  
$I_C = I_2$  
$I_A = I_1 - I_2$ (if $I_1$ flows down and $I_2$ flows up through the middle branch)

Let's apply Kirchhoff's Voltage Law (KVL) to the two independent loops.

**Loop 1 (Left Loop: M-9V-B-N-4Ω-M):**  
Starting from M and going clockwise:

-9V + I_1(1\Omega) + I_1(3\Omega) + I_A(4\Omega) + 3V + I_1(2\Omega) = 0

Wait, the diagram shows the 3V battery in the middle branch. Let's redefine currents more carefully.

Let $I_1$ be the current flowing from M through the $3\Omega$ resistor and 9V battery (left branch).  
Let $I_2$ be the current flowing from M through the $5\Omega$ resistor and 11V battery (right branch).  
Let $I_3$ be the current flowing from M through the $4\Omega$ resistor and 3V battery (middle branch).

**Kirchhoff's Current Law (KCL) at node M:**

I_1 + I_2 + I_3 = 0

This implies that currents are flowing out of M, which is fine, we'll get negative values if they flow in. Or, let's assume currents flow **into** M from the batteries and **out** of M into the middle branch.  
Let $I_1$ be current from 9V battery, $I_2$ from 11V battery, $I_3$ through 4Ω resistor.  
KCL at M: $I_1 + I_2 = I_3$ (assuming $I_3$ flows from M to N)

**KVL for Loop 1 (Left loop: 9V battery, $3\Omega$, $1\Omega$, $4\Omega$, 3V battery):**  
Let's assume $I_1$ flows clockwise in the left loop and $I_2$ flows clockwise in the right loop.  
Current through $3\Omega$ and $1\Omega$ (internal resistance of 9V battery) is $I_1$.  
Current through $5\Omega$ and $3\Omega$ (internal resistance of 11V battery) is $I_2$.  
Current through $4\Omega$ and $2\Omega$ (internal resistance of 3V battery) is $I_A$.  
From KCL at M: $I_1 + I_2 = I_A$ (assuming $I_1$ and $I_2$ flow into M, and $I_A$ flows out of M towards N).  
Ammeter A measures $I_A$. Ammeter B measures $I_1$. Ammeter C measures $I_2$.

Let's use mesh analysis with two loop currents, $I_x$ for the left loop (clockwise) and $I_y$ for the right loop (clockwise).

**Loop 1 (Left loop):**  
The resistors are $3\Omega$, $1\Omega$ (internal to 9V battery), $4\Omega$, $2\Omega$ (internal to 3V battery).  
The batteries are 9V and 3V.

-9V + I_x(3\Omega) + I_x(1\Omega) + (I_x - I_y)(4\Omega) + (I_x - I_y)(2\Omega) + 3V = 0

-9 + 4I_x + 6(I_x - I_y) + 3 = 0

-6 + 4I_x + 6I_x - 6I_y = 0

10I_x - 6I_y = 6 \quad (1)


**Loop 2 (Right loop):**  
The resistors are $5\Omega$, $3\Omega$ (internal to 11V battery), $4\Omega$, $2\Omega$ (internal to 3V battery).  
The batteries are 11V and 3V.

-11V + I_y(5\Omega) + I_y(3\Omega) + (I_y - I_x)(4\Omega) + (I_y - I_x)(2\Omega) + 3V = 0

-11 + 8I_y + 6(I_y - I_x) + 3 = 0

-8 + 8I_y + 6I_y - 6I_x = 0

-6I_x + 14I_y = 8 \quad (2)


Now we have a system of two linear equations:
1) $10I_x - 6I_y = 6$
2) $-6I_x + 14I_y = 8$

Step 1: Solve the system of equations for $I_x$ and $I_y$.  
From (1), divide by 2: $5I_x - 3I_y = 3 \implies 3I_y = 5I_x - 3 \implies I_y = \frac{5I_x - 3}{3}$  
Substitute $I_y$ into (2):

-6I_x + 14\left(\frac{5I_x - 3}{3}\right) = 8

Multiply by 3 to clear the denominator:

-18I_x + 14(5I_x - 3) = 24

-18I_x + 70I_x - 42 = 24

52I_x = 66

I_x = \frac{66}{52} = \frac{33}{26} , A \approx 1.269 , A


Now find $I_y$:

I_y = \frac{5\left(\frac{33}{26}\right) - 3}{3} = \frac{\frac{165}{26} - \frac{78}{26}}{3} = \frac{\frac{87}{26}}{3} = \frac{87}{78} = \frac{29}{26} , A \approx 1.115 , A


**i) Reading of the meters A, B and C**

*   Ammeter B measures $I_x$.
    $$
    I_B = I_x = \frac{33}{26} \, A \approx 1.27 \, A
    $$
*   Ammeter C measures $I_y$.
    $$
    I_C = I_y = \frac{29}{26} \, A \approx 1.12 \, A
    $$
*   Ammeter A measures the current through the middle branch ($4\Omega$ resistor and 3V battery). This current is $I_A = I_x - I_y$ (assuming $I_x$ flows down and $I_y$ flows up, so the net current is $I_x - I_y$ downwards).
    $$
    I_A = I_x - I_y = \frac{33}{26} \, A - \frac{29}{26} \, A = \frac{4}{26} \, A = \frac{2}{13} \, A \approx 0.154 \, A
    $$
    Since $I_A$ is positive, the current flows in the direction of $I_x$ (downwards from M to N).

Readings:  
Ammeter A: $\boxed{\frac{2}{13} \, A \approx 0.154 \, A}$  
Ammeter B: $\boxed{\frac{33}{26} \, A \approx 1.27 \, A}$  
Ammeter C: $\boxed{\frac{29}{26} \, A \approx 1.12 \, A}$

**ii) Potential difference between M and N**

The potential difference between M and N ($V_{MN}$) can be calculated by taking a path from M to N through the middle branch.

V_{MN} = I_A (4\Omega) + I_A (2\Omega) + 3V

V_{MN} = I_A (4\Omega + 2\Omega) + 3V

V_{MN} = \frac{2}{13} , A (6 , \Omega) + 3 , V

V_{MN} = \frac{12}{13} , V + 3 , V = \frac{12 + 39}{13} , V = \frac{51}{13} , V

V_{MN} \approx 3.923 , V

Alternatively, using the left branch (from M to N):

V_{MN} = 9V - I_B (3\Omega) - I_B (1\Omega)

V_{MN} = 9V - I_x (4\Omega)

V_{MN} = 9V - \frac{33}{26} , A (4 , \Omega) = 9 - \frac{132}{26} = 9 - \frac{66}{13} = \frac{117 - 66}{13} = \frac{51}{13} , V

This matches.

The potential difference between M and N is $\boxed{\frac{51}{13} \, V \approx 3.92 \, V}$.

**iii) Power dissipated as heat in the $5\Omega$ resistor**

The power dissipated in a resistor is given by $P = I^2 R$.  
The current flowing through the $5\Omega$ resistor is $I_y$.

P_{5\Omega} = I_y^2 \times 5\Omega

P_{5\Omega} = \left(\frac{29}{26} , A\right)^2 \times 5 , \Omega

P_{5\Omega} = \frac{841}{676} \times 5 , W

P_{5\Omega} = \frac{4205}{676} , W

P_{5\Omega} \approx 6.22 , W


The power dissipated as heat in the $5\Omega$ resistor is $\boxed{\frac{4205}{676} \, W \approx 6.22 \, W}$.

Dernière correction gratuite aujourd'hui — reviens demain, ou tape /upgrade pour l'illimité.

Define diffraction.

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