Inapakia....
|
Bonyeza na Ushikilie kuvuta kuzunguka |
|||
|
Bonyeza Hapa Kufunga |
|||
Swali 1 Ripoti
You are provided with a glass block, plane mirror, and optical pins.
(b)i. Explain the term refractive index and give a mathematical expression for it in terms of wavelength.
ii. State the conditions necessary for total internal reflection to occur for a given pair of media.
The block is traced as ABCD, the width is measured as \(W = 5.0\ \text{cm}\). For each angle of incidence the emergent ray is fixed by no-parallax pins \(P_3\) and \(P_4\), and the angles \(\theta\) and \(e\) together with the lateral displacement \(d\) are measured directly from the traces. The full set of readings and the derived quantities \(m = \sin e\) and \(n = \cos\!\left(\dfrac{\theta}{2}\right)\) are tabulated below.
| \(i/^{\circ}\) | \(\theta/^{\circ}\) | \(e/^{\circ}\) | \(d/\text{cm}\) | \(m = \sin e\) | \(n = \cos\left(\frac{\theta}{2}\right)\) |
|---|---|---|---|---|---|
| 10 | 10.4 | 10.0 | 3.00 | 0.174 | 0.996 |
| 20 | 19.0 | 20.4 | 3.90 | 0.349 | 0.986 |
| 30 | 20.0 | 30.0 | 6.00 | 0.500 | 0.985 |
| 40 | 30.0 | 40.0 | 7.00 | 0.643 | 0.966 |
| 50 | 30.0 | 50.0 | 7.50 | 0.766 | 0.966 |
Sample evaluations: for \(i = 30^{\circ}\), \(m = \sin 30.0^{\circ} = 0.500\) and \(n = \cos\!\left(\tfrac{20.0^{\circ}}{2}\right) = \cos 10.0^{\circ} = 0.985\). For \(i = 50^{\circ}\), \(m = \sin 50.0^{\circ} = 0.766\) and \(n = \cos 15.0^{\circ} = 0.966\).
Two points on the line of best fit are \((n_1, m_1) = (0.996,\ 0.174)\) and \((n_2, m_2) = (0.966,\ 0.766)\).
\[ s = \frac{m_2 - m_1}{n_2 - n_1} = \frac{0.766 - 0.174}{0.966 - 0.996} = \frac{0.592}{-0.030} = -19.7 \]The slope is \(s = -19.7\) (magnitude \(19.7\)).
With \(W = 5.0\ \text{cm}\) and \(s = -19.7\):
\[ q = 2Ws = 2 \times 5.0\ \text{cm} \times (-19.7) = -197\ \text{cm} \]Hence \(|q| = 197\ \text{cm}\).
The refractive index is the ratio of the velocity of light in air (vacuum) to the velocity of light in a material medium as light waves pass from air into the medium. In terms of wavelength, since the frequency is unchanged on refraction,
\[ n = \frac{\lambda_{1}}{\lambda_{2}} \]where \(\lambda_{1}\) is the wavelength in air, \(\lambda_{2}\) is the wavelength in the material, and \(n\) is the refractive index of the material.
Maelezo ya Majibu
The block is traced as ABCD, the width is measured as \(W = 5.0\ \text{cm}\). For each angle of incidence the emergent ray is fixed by no-parallax pins \(P_3\) and \(P_4\), and the angles \(\theta\) and \(e\) together with the lateral displacement \(d\) are measured directly from the traces. The full set of readings and the derived quantities \(m = \sin e\) and \(n = \cos\!\left(\dfrac{\theta}{2}\right)\) are tabulated below.
| \(i/^{\circ}\) | \(\theta/^{\circ}\) | \(e/^{\circ}\) | \(d/\text{cm}\) | \(m = \sin e\) | \(n = \cos\left(\frac{\theta}{2}\right)\) |
|---|---|---|---|---|---|
| 10 | 10.4 | 10.0 | 3.00 | 0.174 | 0.996 |
| 20 | 19.0 | 20.4 | 3.90 | 0.349 | 0.986 |
| 30 | 20.0 | 30.0 | 6.00 | 0.500 | 0.985 |
| 40 | 30.0 | 40.0 | 7.00 | 0.643 | 0.966 |
| 50 | 30.0 | 50.0 | 7.50 | 0.766 | 0.966 |
Sample evaluations: for \(i = 30^{\circ}\), \(m = \sin 30.0^{\circ} = 0.500\) and \(n = \cos\!\left(\tfrac{20.0^{\circ}}{2}\right) = \cos 10.0^{\circ} = 0.985\). For \(i = 50^{\circ}\), \(m = \sin 50.0^{\circ} = 0.766\) and \(n = \cos 15.0^{\circ} = 0.966\).
Two points on the line of best fit are \((n_1, m_1) = (0.996,\ 0.174)\) and \((n_2, m_2) = (0.966,\ 0.766)\).
\[ s = \frac{m_2 - m_1}{n_2 - n_1} = \frac{0.766 - 0.174}{0.966 - 0.996} = \frac{0.592}{-0.030} = -19.7 \]The slope is \(s = -19.7\) (magnitude \(19.7\)).
With \(W = 5.0\ \text{cm}\) and \(s = -19.7\):
\[ q = 2Ws = 2 \times 5.0\ \text{cm} \times (-19.7) = -197\ \text{cm} \]Hence \(|q| = 197\ \text{cm}\).
The refractive index is the ratio of the velocity of light in air (vacuum) to the velocity of light in a material medium as light waves pass from air into the medium. In terms of wavelength, since the frequency is unchanged on refraction,
\[ n = \frac{\lambda_{1}}{\lambda_{2}} \]where \(\lambda_{1}\) is the wavelength in air, \(\lambda_{2}\) is the wavelength in the material, and \(n\) is the refractive index of the material.
Swali 2 Ripoti
You are provided with a voltmeter V, a chemical cell/ battery E; two standard resistors R, and R a potentiometer a key K a jockey, and other necessary materials.
(b)i. State the two devices in which ohm's law does not apply.
ii. A current of 1 A is supplied to two resistors of resistance 2\(\Omega\) and 3\(\Omega\) connected in parallel. Calculate the current in each resistor.
The circuit is connected as shown, with the cell E and key K driving the potentiometer wire AB, and the voltmeter V connected across the length AC = x.
With the key K closed, the jockey is placed on the wire at C so that AC = x, and the voltmeter reading V is taken for each length. The reciprocals \(x^{-1}\) and \(V^{-1}\) are then evaluated.
| \(x\,/\,\text{cm}\) | \(V\,/\,\text{V}\) | \(x^{-1}\,/\,\text{cm}^{-1}\) | \(V^{-1}\,/\,\text{V}^{-1}\) |
|---|---|---|---|
| 20 | 0.15 | 0.050 | 6.667 |
| 30 | 0.20 | 0.033 | 5.000 |
| 40 | 0.30 | 0.025 | 3.333 |
| 50 | 0.40 | 0.020 | 2.500 |
| 60 | 0.50 | 0.017 | 2.000 |
| 80 | 0.60 | 0.013 | 1.667 |
The graph of \(V^{-1}\) (vertical axis) against \(x^{-1}\) (horizontal axis) is a straight line passing through the origin.
Taking two widely separated points on the line of best fit, \((0.050,\;6.667)\) and \((0.013,\;1.667)\):
\[ s = \frac{V_2^{-1}-V_1^{-1}}{x_2^{-1}-x_1^{-1}} = \frac{6.667-1.667}{0.050-0.013} = \frac{5.000}{0.037} = 135.1\ \text{cm}\,\text{V}^{-1} \]The line of best fit passes through the origin, so the intercept on the vertical axis is
\[ c = 0 \]A total current of \(1\ \text{A}\) is supplied to \(2\ \Omega\) and \(3\ \Omega\) in parallel. The current divides in the inverse ratio of the resistances:
\[ I_{2\Omega} = 1 \times \frac{3}{2+3} = \frac{3}{5} = 0.6\ \text{A} \] \[ I_{3\Omega} = 1 \times \frac{2}{2+3} = \frac{2}{5} = 0.4\ \text{A} \]Therefore \(0.6\ \text{A}\) flows through the \(2\ \Omega\) resistor and \(0.4\ \text{A}\) through the \(3\ \Omega\) resistor.
Maelezo ya Majibu
The circuit is connected as shown, with the cell E and key K driving the potentiometer wire AB, and the voltmeter V connected across the length AC = x.
With the key K closed, the jockey is placed on the wire at C so that AC = x, and the voltmeter reading V is taken for each length. The reciprocals \(x^{-1}\) and \(V^{-1}\) are then evaluated.
| \(x\,/\,\text{cm}\) | \(V\,/\,\text{V}\) | \(x^{-1}\,/\,\text{cm}^{-1}\) | \(V^{-1}\,/\,\text{V}^{-1}\) |
|---|---|---|---|
| 20 | 0.15 | 0.050 | 6.667 |
| 30 | 0.20 | 0.033 | 5.000 |
| 40 | 0.30 | 0.025 | 3.333 |
| 50 | 0.40 | 0.020 | 2.500 |
| 60 | 0.50 | 0.017 | 2.000 |
| 80 | 0.60 | 0.013 | 1.667 |
The graph of \(V^{-1}\) (vertical axis) against \(x^{-1}\) (horizontal axis) is a straight line passing through the origin.
Taking two widely separated points on the line of best fit, \((0.050,\;6.667)\) and \((0.013,\;1.667)\):
\[ s = \frac{V_2^{-1}-V_1^{-1}}{x_2^{-1}-x_1^{-1}} = \frac{6.667-1.667}{0.050-0.013} = \frac{5.000}{0.037} = 135.1\ \text{cm}\,\text{V}^{-1} \]The line of best fit passes through the origin, so the intercept on the vertical axis is
\[ c = 0 \]A total current of \(1\ \text{A}\) is supplied to \(2\ \Omega\) and \(3\ \Omega\) in parallel. The current divides in the inverse ratio of the resistances:
\[ I_{2\Omega} = 1 \times \frac{3}{2+3} = \frac{3}{5} = 0.6\ \text{A} \] \[ I_{3\Omega} = 1 \times \frac{2}{2+3} = \frac{2}{5} = 0.4\ \text{A} \]Therefore \(0.6\ \text{A}\) flows through the \(2\ \Omega\) resistor and \(0.4\ \text{A}\) through the \(3\ \Omega\) resistor.
Swali 3 Ripoti
You are provided with three retort stands, a pendulum bob, a drawing board, a stopwatch, and other necessary apparatus. Using the diagram above as a guide, carry out the following instructions.
(b)i. Distinguish between the period and frequency of oscillation of a simple pendulum.
ii. Differentiate between oscillatory and rotational motions.
The drawing paper is fixed to the vertical board. The line RP is drawn as the rest position of the string, with P marked at the centre of the bob at rest. When the bob is displaced to a new position P¹, the perpendicular distance d of P¹ from the line RP and the vertical height h of P¹ above P are measured as shown below.
For each of the five positions the readings are taken and \(d^{2}\) and \(G=\dfrac{d^{2}}{h}\) are evaluated.
| S/N | d / cm | d² / cm² | h / cm | G = d²/h / cm |
|---|---|---|---|---|
| P₁ | 3.30 | 10.89 | 3.00 | 3.63 |
| P₂ | 5.50 | 30.25 | 2.80 | 10.80 |
| P₃ | 7.60 | 57.76 | 2.60 | 22.22 |
| P₄ | 9.60 | 92.16 | 2.40 | 38.40 |
| P₅ | 10.50 | 110.25 | 2.20 | 50.11 |
Sample evaluations: \(G_{1}=\dfrac{10.89}{3.00}=3.63\,\text{cm}\), and \(G_{5}=\dfrac{110.25}{2.20}=50.11\,\text{cm}\).
With the board removed, the bob is set oscillating through a small amplitude. Time for 20 oscillations, \(t = 21.76\,\text{s}\), \(n = 20\).
\[T=\frac{t}{n}=\frac{21.76}{20}=1.088\,\text{s}\]G is plotted on the vertical axis against h on the horizontal axis, both axes starting from the origin.
The best-fit line cuts the horizontal (h) axis where \(G=0\). Reading from the graph, the intercept on the horizontal axis is
\[I \approx 3.0\,\text{cm}\]| Period (T) | Frequency (f) |
|---|---|
| The time taken to complete one complete oscillation (one to-and-fro cycle). Its SI unit is the second (s). | The number of complete oscillations made in one second. Its SI unit is the hertz (Hz). |
They are reciprocals: \(f=\dfrac{1}{T}\).
| Oscillatory motion | Rotational motion |
|---|---|
| A body moves to and fro (back and forth) about a fixed mean position along a path, repeating the motion at regular intervals. Examples: a swinging pendulum, a vibrating string, a loaded test tube bobbing in water, a vibrating tuning fork. | A body turns (spins) about a fixed axis passing through or near the body, every particle moving in a circle about that axis. Examples: the blades of a fan in motion, the Earth spinning about its axis, a spinning wheel. |
Maelezo ya Majibu
The drawing paper is fixed to the vertical board. The line RP is drawn as the rest position of the string, with P marked at the centre of the bob at rest. When the bob is displaced to a new position P¹, the perpendicular distance d of P¹ from the line RP and the vertical height h of P¹ above P are measured as shown below.
For each of the five positions the readings are taken and \(d^{2}\) and \(G=\dfrac{d^{2}}{h}\) are evaluated.
| S/N | d / cm | d² / cm² | h / cm | G = d²/h / cm |
|---|---|---|---|---|
| P₁ | 3.30 | 10.89 | 3.00 | 3.63 |
| P₂ | 5.50 | 30.25 | 2.80 | 10.80 |
| P₃ | 7.60 | 57.76 | 2.60 | 22.22 |
| P₄ | 9.60 | 92.16 | 2.40 | 38.40 |
| P₅ | 10.50 | 110.25 | 2.20 | 50.11 |
Sample evaluations: \(G_{1}=\dfrac{10.89}{3.00}=3.63\,\text{cm}\), and \(G_{5}=\dfrac{110.25}{2.20}=50.11\,\text{cm}\).
With the board removed, the bob is set oscillating through a small amplitude. Time for 20 oscillations, \(t = 21.76\,\text{s}\), \(n = 20\).
\[T=\frac{t}{n}=\frac{21.76}{20}=1.088\,\text{s}\]G is plotted on the vertical axis against h on the horizontal axis, both axes starting from the origin.
The best-fit line cuts the horizontal (h) axis where \(G=0\). Reading from the graph, the intercept on the horizontal axis is
\[I \approx 3.0\,\text{cm}\]| Period (T) | Frequency (f) |
|---|---|
| The time taken to complete one complete oscillation (one to-and-fro cycle). Its SI unit is the second (s). | The number of complete oscillations made in one second. Its SI unit is the hertz (Hz). |
They are reciprocals: \(f=\dfrac{1}{T}\).
| Oscillatory motion | Rotational motion |
|---|---|
| A body moves to and fro (back and forth) about a fixed mean position along a path, repeating the motion at regular intervals. Examples: a swinging pendulum, a vibrating string, a loaded test tube bobbing in water, a vibrating tuning fork. | A body turns (spins) about a fixed axis passing through or near the body, every particle moving in a circle about that axis. Examples: the blades of a fan in motion, the Earth spinning about its axis, a spinning wheel. |
Je, ungependa kuendelea na hatua hii?