(C) PUE KEAkea.kar.nic.in/vikasana/physics_puc/e1_answers.pdf · Electrostastics 1 2 1 T/P 1 C...
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(C) PUE KEA
(C) PUE KEA
TOPIC 1 mark
2 mark
4 mark
5 mark theory
5 mark problem
TOTAL 10 14 4 8 4
Geometrical Optics
1 1 1 1 C
Physical Optics 3 3 1 PC 1
Electrostastics 1 2 1 T/P 1 C
Atomic Physics 3 4 1 T/P 4 1 C
Current Electricity
2 4 1 T/P 2 1 C
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n1 sin i = n2 sin r
Laws of refraction
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LATERAL SHIFT
When a ray of light is incident
obliquely on a parallel sided glass
slab the emergent ray shifts
laterally . The perpendicular
distance between the direction of
the incident ray and emergent ray
is called “lateral shift’’.
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N`
P
A
i
r i-r
B
D CN
T
S
R
t
Q
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ABCD=Principal section
of a glass slab
PQ = incident ray
QR = refracted ray
RS = emergent ray
NN`= normal
RT = lateral shift
BC = t = thickness of
the slab
i = angle of incidence
r = angle of refraction
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Consider ▲NQR,cos r =
From the ▲QRT,sin(i-r)=
QN
QR
QR=QN
cos r
RT
QR
RT cos r
QN
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N`
i
rN
It R
PQ
O
i
r
NORMAL SHIFT
The apparent shift in the position of an
object placed in one medium and viewed
along the normal, from the other medium.
S
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OP = incident ray
PQ = refracted ray produced
backwards meet ON at I
NN`= normal
OI = normal shift
RPO = i = angle of incidence = NPO
QPS = r = angle of refraction = NIP
t = thickness of the optical medium
From the ▲ NPO
From the ▲ NPI
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n1
sin i = n2
sin r
Here n1
= n n2
= 1
P is a point which very
close to N
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TOTAL INTERNAL REFLECTION
denser medium
i Ci>C
r 900
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A
i1 i1r1 r2
dR
S
Z
M
N
Q
P
B C
REFRACTION THROUGH A PRISM
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ABC = principal section
of the prism
BC = Base
A = Angle of the prism
PQ = Incident ray
QR = Refracted ray
RS = Emergent ray
ZMS = d = Angle of deviation
i1
= Angle of incidence
r1
= Angle of refraction
r2
= Angle of incidence at the
face AC
i2
= Angle of emergence
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Consider the Quadrilateral AQNR
In the ΔQNR
A + QNR = 1800
r1
+ r2
+ QNR = 1800
equating the above two equations, we have
A + QNR = r1
+ r2
+ QNR
A = r1
+ r2
---------------------(1)
d = (i1
– r1) + (i
2– r
2)
d = (i1
+ i2) – (r
1+ r
2) = i
1+ i
2- A (from 1)
d + A = i1
+ i2
---------------------(2)
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(1) A = 2r or r = A/2
(2) D + A = 2i or i = (A+D)/2
D
i
d
i
at minimum deviation
i1
= i2
= i
r1
= r2
= r
d = D
i1 i2
d
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REFRACTION THROUGH
CURVED SURFACES
Curved surface: It is a part of a sphere.
Principal axis: The horizontal line passing through the centre of the curved surface.
Pole(P): Centre of the curved surface.
Centre of curvature(C): It is a point on the principal axis which is the centre of the sphere in which curved surface is a part.
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Radius of curvature(R): It is the
distance between centre of curvature and the
pole
Principal focus(F): It is a point on the
principal axis where parallel rays from infinity
converge or appear to diverge from that point
Focal length (f): It is the distance
between the principal focus and the pole.
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Relation between n,u,v&R
C O I P
M
ϴ
i
rN
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OM = incident ray
MN = refracted ray
MI = refracted ray produced
backwards
CP = radius of curvature (R)
OP = object distance (u)
IP = image distance (v)
i = angle of incidence
r = angle of refraction
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Applying sine rule to the ▲s CMO & CMI
Dividing equation (2) by equation (1)
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= X
=
=
We know that n1
sin i = n2
sin r
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Dividing the above equation through out by uvR
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M is a point which is very
close to P, therefore
IM ≈ IP = -v (because image
is virtual)
OM ≈ OP = u
CO = CP – OP = R - u
CI = CP - IP = R + v
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Lens maker’s formula
u v I
A
C
B DP
O I`
V`
E F
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OE = incident ray
FI = refracted ray
O = point object
I` = virtual object for the surface ADC
OP = u =object distance
R1
,R2
=radii of curvatures of the
surfaces ABC and ADC
PI =v=image distance
P I` =v` =image distance for the first
surface
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For the refraction through surface ABC,
For the refraction through surface ADC,
Negative sign indicates that the
object is virtual
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Adding (1) and (2)
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Two thin lenses in contact
L1 L2
O I I`
u vV`
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Let us consider two thin lenses in
contact let f1
and f2
be their focal
lengths
let u=object distance
v =image distance
v`=virtual image distance
here I`is the image formed
For refraction through first lens,
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For refraction through second lens,
here I`serves as virtual object for the
second lens.
Adding equations (1) and (2)
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let F be the effective focal length
of the combination.
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Questions carrying ONE mark each
1. Define critical angle.
2. Mention an application of total internal
reflection.
3. What is dispersion of light?
4. Write the expression for the deviation produced
by a thin prism.
5. Define dispersive power of the material of a prism.
6. What is angular dispersion?
7. Write the condition for dispersion without deviation.
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14. A thin convex lens of focal length 0.1m & a thin
concave lens of focal length 0.1m are placed co-
axially in contact. What is the net power of the
combination?
8. Define power of a lens.
9. What is magnification?
10. Write the relation between power and focal
length of a lens.
11. Write the expression for the equivalent focal
length of two thin lenses separated by a distance.
12. What is a thin prism?
13. What is the lateral shift produced by a parallel
sided glass slab for grazing incidence of light?
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Questions carrying two marks
1.Mention the conditions for total internal
reflection to occur.
2.Arrive at the relation between critical angle and
refractive index.
•the light ray should travel from denser medium to rarer
medium
•the angle of incidence must be greater than critical
angle
c
Here n1=n, i=c, n
2=1 & r=90
0
n sinc=1 sin 900
n sinc=1
n=
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3.Write the ray diagram showing the
experimental arrangement for pure spectrum.
S
L1L2
R
V
P
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6.What is an impure spectrum? give an example.
5.What is a pure spectrum? give an
example.
If the constituent colours are seen distinctly in a
spectrum, it is called pure spectrum.
Example: Spectrum produced in the laboratory
A spectrum in which the constituent colours overlap
and are not seen distinctly is called impure spectrum.
Example: Rainbow
4.What is an optical fibre? on what principle
does it work?
An optical fibre is a thin transparent fibre which
transmits light along any desired path. It works on
the princple of TIR.
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8.Calculate the angle of deviation produced by a
thin prism of refracting angle5˚ and refractive
index 1.5.
7.Calculate the refractive index of a pair of media
given critical angle as 40˚.
given C=40˚ ,n=1
d=?,A=5˚,n=1.5 d=(n-1)A
=(1.5-1)X5˚
=0.5X5˚=2.5˚
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9.An object is kept at a distance of 0.12m
from a convex lens forms an image at a
distance of 0.18m.calculate the
magnification produced.
u=0.12m,v=0.18m, m=?
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10.Two thin convex lenses of focal
lengths 0.15m & 0.2m are separated by
a distance of 0.6m. Find the effective
focal length of the combination.
given:f1=0.15,f
2=0.2,d=0.06,F=?
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Question carrying FIVE mark each
1. What is lateral shift? Derive an expression for
lateral shift produced by a parallel sided glass
slab for oblique incidence.
2. What is normal shift? Derive the expression for
normal shift produced when an object in a
denser medium is viewed normally through air.
3. Derive the expression for refractive index in
terms of the angle of the prism and angle of
minimum deviation.
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4.Derive the relation connecting n,u,v & R for
refraction at a spherical surface concave
towards a point object in a denser
medium.
5.Derive lens maker’s formula.
6.Derive the expression for the effective
focal length of two thin lenses in
contact.
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•Read the given problem twice
•Write the given data
•Write a relevant formula/e connecting the given data
•Substitute and simplify
•Write the result with S.I unit
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List of formulae
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March 2010 : Find the real position of an air
bubble in a glass cube of side 0.3 m if the bubble
appears to be at a distance of 0.05 m from one face
and at 0.15 m from the opposite face.
Solution:
Here apparent distance from both the faces are given.
A.D1
= 0.05 m
A.D2
= 0.15 m
Let ‘x’ be the real distance from the first
face.
Real distance from the opposite face is
(0.3-x) m
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We know that
Therefore the real position of the air
bubble in the glass cube is 0.1 m from
the first face or 0.2 m from the
opposite face.
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March 2008: Focal length of convex lens is
0.1 m. A liquid lens is formed between a plane
surface and one face of this lens of radius of
curvature 0.12m. The converging combination
formed is found to have a focal length 0.18 m.
Calculate the refractive index of liquid.
Solution:
Given F=0.18 m, f1
= 0.1m, R1=0.12m, n=?
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Here radius of curvature of the plane surface
is infinity
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March 2011: A concave lens when placed in
air has a focal length of 0.4 m and has a refractive
index of 3/2. What will be its focal length when
immersed in water of refractive index 4/3.
Solution:
Given: f=0.4 m, n of glass=3/2, f of lens when
immersed in water=? R.I of water 4/3
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Simplify and arrive at
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March 2006: A small air bubble is found
0.04 m inside the glass sphere of refractive
index 1.5 When viewed normally from outside it
is seen at a depth of 0.03 m. Calculate the
radius of curvature of the glass sphere.
Solution:
Given
u=0.04m, n=1.5, v=0.03m, R?
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R=0.119m