Activity Sec - B
Activity Sec - B
ACTI VITY BI
AIM
Carbon resistor
Dr Diodes
LED
1000 uF
6V
0.47 uf
Paper Electrolytic
Mica CAPACITORS
4. If the multimeter shows an equal deflection in both directions, then the component is aresisio
light from h.
5. 1f amultimeter shows deflection in one direction without any emission of
diode
component and no deflection in the opposite direction, then the component is a
light from h
6. If the multimeter shows deflection in one direction alongwith the emission of
is an LED
component and no deflection in the opposite direction, then the component
7. If the multimeter does not show any deflection on connecting its probes either way to a
capacitor is laroe
component, then the component is acapacitor. But if the capacitance of the
the multimeter will show a large deflection initially which gradually decreases to zero.
8. Record all your observations in a tabular form.
OBSERVATIONS
Table B1 : State of conduction of each component
RESULT
From the mixed grouping of components, the components marked A, B, Cand Dhave been
identified as resistor, diode, LED and capacitor respectively.
PRECAUTIONS
1. Whilechecking the conduction state of any component, clean its leads properly.
2. Use the selector switch of the multimeter in resistance mode with highest range option.
3. While testing any component, avoid touching the metal end of either of multimeter probe.
Body resistance in parallel with the component resistance may create confusion about the
conduction state of the component.
VIVA VOCE
1. What is aresistor ? 5. How do the conduction states of an ordinary diode
Any material that has some resistance is called a and an LED differ ?
resistor. An ordinary diode conducts in forward biasing
2. What is a linear resistor ? without any emission of light while an LED conducts
A linear resistor is one which obeys Ohm's law or for in forward direction with emission of light. Both do
not conduct in reverse biasing.
which V-lgraph is a straight line passing through the
origin. 6. How does a capacitor behave towards dc?
3. What is a non-ohmic device ? A capacitor does not conduct dc. But a capacitor of
large capacitance shows an initial deflection in the
A device which does not obey Ohm's law is called a
ammeter which decays to zero quickly. This is due to
non-ohmic device. Semiconductor diodes, LEDS, etc;
are non-ohmic devices. charging of the capacitor.
4. Does an ohmic resistor conduct equally for both 7. How does a capacitor behave towards ac ?
forward and reverse biasing ? A capacitor conducts ac because its capacitive
Yes, an ohmic resistor conducts equally when current is is finite against ac.
passed in one direction and then in opposite direction. 2C)
182
Section B: ACTIVITIES
ACTI VIT Y B2
AIM
Use of multimeter to see the unidirectional flow of current in case of a diode
and an
LED and check whether a given electronic component (e.g., diode) is in working order.
APPARATUS AND MATERIAL REQUIRED
A multimeter, a diode, LED, sand paper.
THEORY
1. To check the unidirectional flow of current through a junction diode/LED. When a
junction diode/LED is forward biased, a substantial current ( few mA) willflow through it.
When a junction diode/LED is reverse biased, a negligible current ( few A) will flow
through it.
2. To check whether a diode is in working order. A junction diode offers a low resistance (a
few 2 to k) during forward biasing and it offers a very high resistance (r M2) during
reverse biasing. Thus the working of a junction diode can be examined by measuring its
resistance in forward and reverse biased conditions.
WORKING DIAGRAM
Diode R
RE
p-n
+ Red probe
- Black probe
K 6V variable
dc battery
Fig. 2 To check unidirectional flow of current through a diode.
PROCEDURE
(a) To check unidirectional flow of current through a junction diode/LED
1. As shown in Fig. 2, connect the junction diode, a resistance box, a 6 V variable dc battery
and a plug key K in series. Adjust the battery to minimum voltage.
2. Set multimeter in current measuring mode at a suitable range of mA (starting from high
current
current range). Take out a suitable resistance R from the resistance box so that the
K and note the value of urrent
flows within the range chosen. Insert the plug in the key
flowing in the circuit.
3. Increase the forward bias in steps of 0.2 V. Note the current in each case. Beyond a certain
forward bias.
applied voltage, current increases rapidly with the increase in
4. Reverse the terminals of the junction diode so that its pend is at lower potential and rend is
at hicher notential. Observe the current in multimeter on uA scale. Negligible current
reading will indicate the unidirectional teature of the diode. Increase in reverse bias will
show negligible change in reverse current.
183
LAB MANUAL PHYSICS-XII
5. Replace the diode by LED and repeat steps 1to 4. It will be seen that LED allows the flow
when the
current only when it is forward biased. LED starts glowing applied
exceeds 1.2 V. Initially, it glows faintly. lts glow becomes brighter and brighter as the voltage
is gradually increased. voltage
(b) To check whether the junction diode is in working order or not
6. Set the multimeter in resistance measuring mode.
7. Touch the two probes of the multimeter to the two end terminals of the junction diode. Note
the diode resistance. Reverse the diode connections. Again note the diode resistance. If the
diode resistance is low in one case and high in the other case or Vice versa, then the diode is
in working order.
8. If the diode resistance is low both during forward and reverse biasings, then the diod :
short-circuited. If the diode resistance is very high both during forward and reu
biasings, then the diode junction is discontinuous or open. In both of these situations #he
diode is not in working order. Record the observations in a tabular form.
OBSERVATIONS
A.For unidirectionalflow of current through the junction diode
Table B2: When the junction diode is connected in forward bias
Forward bias voltage Forward current
S.No.
(V) (mA)
1
2
3.
4
5
5.
3.
4
5
184
Section B: ACTIVITIES
185
LAB MANUAL PHYSICS-X|
ACTIVIT Y
B3
AIM
To study the effect of intensity of light (by varying distance of the source) on
an LDR.
Metal
-Photoconductive
material over
contact
top surface
Ceramic
substrate
+-Wire
terminal
(a) (b)
When light of certain minimum frequency falls on the LDR, the absorbed photons give bound
electrons enough energy to jump to the conduction band where they are free to conduct
electricity. This increases the conductance or lowers the resistance of the device. Atypical LDR
has a high resistance of several M2 in total darkness and just a few hundred ohms in
bright light.
186
Seclion B: ACTIVITIES
CIRCUIT DIAGRAM
To ac
mains
3 W LED
47 SQ K mA
LDR
Iron
stand
12 V
Range of milliammeter =0 to mA
2 15
3 20
4. 25
5 30
RESULT
1. As distance of LDR from the source increases, intensity of light decreases and resistance of
LDR increases.
PRECAUTIONS
1. All the connections should be neat and tight.
2. LDR must be placed normally to the light source so that angle of incidence of light rays
remains same throughout the experiment.
3. A suitable protective resistance must be connected in series with LDR to prevent it from
damage.
4. Resistance of LDR for background illumination must be taken into account.
VIVA VOCE
1, What is an LDR ? bands of the photosensitive material. These electrons
A light dependent resistor is a device whose become free to conduct electricity. This increases the
resistance decreases with the increase in intensity of conductance or decreases the resistance.
light incident upon it. 5. How does the intensity of light vary with the distance
from the source?
2. Name some mnaterials used for nmaking LDRs.
Semiconductor materials like CdS, PbS, CdSe, PbSe Intensity of light is inversely proportional to the
and InSb. square ofthe distance from the source.
6. How are LDRs different from photodiodes and
3. How does the resistance of an LDR change with the phototransistors ?
light intensity ? Even though LDRs are made from semiconductor
At low light intensities, an LDR has a high resistance. materials, yet they are different from photodiodes
As the intensity of incident light increases, the and phototransistors because they are simply passive
resistance of an LDR decreases.
devices and do not have p-n junctions.
4. Why does the resistance of an LDR decrease with the 7. Name some
increase in intensity of incident light ? applications of LDRs.
As the intensity of incident light increases, more and LDRs are used as light sensors in alarm clocks, street
more electrons are knocked off from the valence lights, light intensity meters and burglar alarm
circuits.
188
Section B: ACTIVITIES
ACTIVIT Y B4
AIM
Air
B
Glass
Thus the emergent ray CD is parallel to the incident ray AB But the emergent ray gets laterally
displaced with respect to the incident ray. The perpendicular distance between the incident and
emergent rays, when light is incident obliquely on arefracting slab with parallel faces, is called laterial shift
or lateral displacement.
It is given by
t cosi
d=
COS I
sin(i -r) =tsin i 1 (u-sin i)2
=t sin 90= t
Clearly, dmax
The lateral shift produced by a glass slab increases with
() the increase in thickness t of glass slab,
(ii) the increase in the value of angle of incidence i, and
(ii) the increase in refractive index å of the glass slab.
189
LAB MANUAL PHYSICS-XII
RAY DIAGRAMS
A N A, N N3
P
i=40° i=40o
P
i=40°
p
B B:M
R
t
D,
R
D
(a (b) (c)
500 As P, 60
P
P
B
Ds
(d .(e)
Fig. 6 Lateral displacements for different i and t.
PROCEDURE
1. Fix a white sheet of paper on the drawing board with the help of cello-tapeor drawing pins.
2. Place the glass slab breadthwise on the white sheet [Fig. 6(a)]. Mark its boundary PQRS with a
sharp pencil.
3. Remove the glass slab. Take a point B, on face AB Draw normal B, N, on PQ. With the help of a
protractor, draw an inident ray A, B, making an angle of incidence of 40° with the normal B, N,.
4. Again, place the glass slab within its boundary PQRS. Fix two alpins P, and P, vertically on
the incident ray A, B,, about 8 to 10 cm apart.
5. Looking into the slab from the opposite face SP, position the eye in sucha way that the feet of
pins P and P, appear to be one behind the other. Now fix pins P, and P, vertically in line
with pins P, and P, as viewed through the slab.
6. Remove the alpins and encircle the pin-pricks. Remove the slab andcomplete the path of the
ray of light A, B, C D, Draw perpendicular CE on A, B, produced and measure the length
of C, E,. This gives a measure of lateral displacement d.
7. Place the glass slab lengthwise as shown in Fig. 6(b). Repeat the experiment again for the same
angle of incidence of 40°. Measure the lateral displacement.
8. Place the glass slab thicknesswise as shown in Fig. 6(c). Repeat the experiment again for the
same angle of incidence of 40°, Measure the lateral displacement.
9. By placing the glass slab lengthwise as shown in Figs. 6(d) and (e), repeat the experiment twice
for angles of incidence of 50° and 60. Measure the lateral displacement in each case.
190
Section B ACTIVITIES
10. Measure the length, breadth and thickness of the glass slab
using a ruler. Record all your
observations in a tabular form.
OBSERVATIONS AND CALCULATIONS
Least count of the
protractor = degrees
Least count of the ruler mm cm ; Length of the glass slab, t, Cm
Breadth of the glass slab, t, = Cm ; Thickness of the glass slab, t3 = Cm
3 40° ...
4 t 50°
5 t, 60
RESULT
1. As difference i~ eis small, so i =e When light refracts through a glass
slab, the emergent ray
is parallel to the direction of incident ray.
2. Within the limits of experimental error, from observations 1, 2 and 3 we see that the ratio
d|t =constant. The lateral displacement of the emergent ray is directly proportional to the
thickness of the glass slab (for constant i).
3. From observations 2, 4 and 5, we note the lateral displacement of the
emergent ray increases
with the increase in the angle of incidence i(for constant f).
PRECAUTIONS
1. The boundary of the glass slab should be marked with a sharp pencil.
2. Alpins should be fixed vertically and about 8 to 10cm apart.
3. The feet of the alpins and not their heads should be adjusted in the same straight line.
4. Just after removing an alpin, encircle the pin-prick with a sharp pencil.
5. The angle of incidence should lie between 30° and 60°.
VIVA VOCE
1. What is lateral shift in refraction ? 3. For what angle of incidence, the lateral shift
The sidewise shift in the path of light on emerging produced by a parallel sided glass slab is zero ?
from a refracting medium with parallel faces is called For i= 0° lateral shift is zero.
lateral shift. 4. For what angle of incidence, the lateral shift
2, On what factors does the lateral shift
depend ? produced by a parallel sided glass slab is maximum ?
Lateral shift depernds on arngle of incidence, the For i =90°, lateral shift is maximum and is equal to
refractive index and thickness of the refracting the thickness of the glass slab.
medium.
191
Dec 3