Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
REFRACTION THROUGH SPHERICAL
SURFACES
1 INTIRODUCTION
A source of light may be considered to give out energy
either as a stream of particles called photons or as a continuous
stream of energy along a ray or as wave motion in the medium.
The concept of light particles such as photons is of importance
in the study of the origin of spectra. It is also useful in the study
of the interaction of light with matter as in the case of
photoelectric effect.
However, as far as propagation of light through a medium is
concerned it is more convenient to suppose that light energy is
propagated as wave motion. From the point of view of wave
motion, a ray can be defined as an imaginary line drawn in the
direction in which the wave is travelling. In the case of
reflection and refraction, the light energy is propagated along
the wave normals. Defining rays as wave normals, it is easy to
conceive that light behaves as a stream of energy propagated
along the direction of the rays. The rays are directed outwards
from the source and they obey the laws of reflection and
refraction. Therefore, the ray concept is of great importance in
the study of geometrical optics.
The processes involved in geometrical optics are only
reflection and refraction. The section on geometrical optics
refers only to the study of light rays undergoing reflection and
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
refraction in optical instruments. Here, one is mainly concerned
with the dominant direction of the ray and other considerations
(e.g. diffraction, interference) are not taken into account. In a
pin hole camera, the geometrical optics deals with the formation
of a well defined image only and does not take into account the
interference and diffraction phenomena.
This is the reason why an optimum size of the pin hole
(0.035 cm) is necessary in a pin hole camera to obtain a well
defined image. Well defined image is not formed when the size
of the pin hole is too large because for each point of the object,
there will be a corresponding patch in the image due to large
number of rays. The overlapping of the patches results in a
blurred image when the size of the hole is comparable to the
wavelength of light, the diffraction effect become prominent
and the ray concept does not hold good.
Thus, the geometrical treatment is sufficient as long as the
surfaces and other discontinuities encountered by the light wave
during its propagation are very large as compared to the
wavelength of light. As long as this condition is satisfied, the
geometrical treatment concerning reflection and refraction is
equally applicable to light waves, ultrasonic waves, earthquake
waves etc.
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
2. REFRACTIVE INDEX AND OPTICAL PATH
The velocity of a monochromatic beam of light in a material
medium is different from that in vacuum. If c and v are the
velocities of light in free space and in a medium respectively,
the ratio c/v is defined as the refractive index of the medium.
The refractive index is denoted by .

c
v
Also c = f where f is the frequency of the vibration. When
light is propagated from one medium to another medium, the
frequency remains unaltered. Consequently, the wavelength is
different in different media. If  and m are the wavelengths in
free space and in the medium, then


m
An optical path is defined as the product of geometrical
distance and the refractive index of the medium. Suppose a ray
travels a distance x in a medium of refractive index , the
optical path is equal to x. In a given time, light travels the same
optical path in different media. Suppose in time t, light travels a
distance X1 in a medium of refractive index 1, and a distance
X2 in a medium of refractive index 2, the optical path,
1 X 1  2 X 2
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
3. FERMAT'S PRINCIPLE OF STATIONARY TIME
In 1658, Fermat enunciated the principle of least time for the
path followed by light radiations. According to this principle,
the path actually taken by a ray of light in passing from one
point to the other is the path of least time. The fundamental laws
of rectilinear propagation, reflection and refraction can be
derived from this principle. However, in some cases it has been
shown that the time taken by light is not minimum but
maximum. Therefore, in the modified form, Format's principle
of least time is known as Fermat's principle of stationary time.
According to this principle, the path taken by a ray of light in
passing from one point to the other is the path of minimum or
maximum time.
(1)
Rectilinear peopagation of light.
As a straight line is the path of least distance between two
points, there fore the time taken by the ray of light along a
straight line is minimum as compared to some other path which
may be curved. Therefore, Fermat's principle shows that light
rays travel along straight line paths in a homogeneous medium.
(2) Derivation of the laws of reflection.
Consider a plane mirror and a plane ABCD normal to it
(Fig. 2.1).
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
Fig. 2-1.
The incident ray AO is reflected along OB. ON is the
normal. The rays, AO, OB and ON lie in the plane ABCD.
Suppose a light ray is reflected from a point P where OP is normal to the plane ABCD. It means the incident ray AP is reflected
along PB. Here, AP>AO and PB>OB. Therefore the path
AP+PB is greater than AO+OB, which is against Fermat's
principle. Therefore, P must coincide with 0 for the path to be
minimum. Hence, AO, OB and ON all lie in the same plane.
Suppose, the angle of incidence is i and the angle of
reflection is r. The point 0 is to be located along CD (Fig. 2.2).
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
Fig. 2-2
Taken AO=u, OB=v, AD=m, BC=n and velocity of light = c
uv
c
OA u
In the  AOD, sec i =
or u = m sec i

AD m
OB V
In the  BOC, sec r =
or v = n sec r

BC n
1

t  ( m sec i  n sec r )
c
Time taken along AOB=t=
(i)
If the point 0 is shifted, there is a slight change in i and r.
Differentiating equation (i)
1
dt  ( m sec i tan i di  n sec r tan r dr )
c
For the time to be minimum
dt = 0

or
1
( m sec i tan i di  n sec r tan r dr )  0
c
m sec i tan i di =- n sec r tan r dr
(ii)
Whatever may be the position of O, DO+OC = constant
But

DO = m tan i and
OC = n tan r
m tan i + n tan r = constant
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
Differentiating;
m sec2 i di+n sec2 r dr=o
or
m sec2 i di = -n sec2 r dr
(iii)
Dividing (ii) by (iii)
i=r
sin i = sin r or
(iv)
Therefore, (i) the angle of incidence is equal to the angle of
reflection and (ii) the incident ray, the reflected ray and the
normal lie in the same plane.
(3)
Derivation of the laws of refraction.
Suppose M1 M2 is the boundary separating the two media of
refractive indices 1 and 2 (Fig. 1.3).
Fig. 2.3
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
c
Velocity of light along AO is
1
and along OB is
c
2
. The
angle of incidence is i and the angle of refraction is r for a ray of
light AOB. Time taken from A to B,
1u
=t=
c

2v

 2 n sec r
c
In the  AOD, u=m sec i
In the  BOC, v=n sec r
=t=
 1 m sec i
c
(i)
c
If the point 0 is shifted, there is slight change in i and r.
Differentiating equation (i)
1
dt  (  1 m sec i tan i di   2 n sec r tan r dr )
c
(ii)
For the time to be minimum,
dt = 0

1 m sec i tan i di +2 n sec r tan r dr=0
1 m sec i tan i di=- 2 n see r tan r dr
or
(iii)
Whatever may be the position of the point 0,
DO + OC = constant.
DO = m tan i
OC = n tan r

m tan i + n tan r = constant.
Differentiating,
m sec2 i di + n sec2 r dr = 0
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Dr M. A. M. EL-Morsy

REFRACTION THROUGH SPHERICAL SURFACES
m sec2 i di = -n sec2 r dr
1 sin i=2 sin r
Dividing (iii) by (iv)
(iv)
(v )
This equation represents Snell's law of refraction. Similar to
reflection, it can be proved in this case also that the incident ray,
the refracted ray and the normal lie in the same plane.
4. MAXIMUM TIME
In some cases it has been observed that a ray of light in passing
between two points takes the path of maximum time. Consider a
spherical reflecting surface AOB with its centre at C and radius
R (Fig. 1.4).
Fig. 2.4.
When an object is placed at P, its image in formed at Q
according to the laws of reflection. The path traversed by light
in this case is POQ. Consider a possible path PDQ.
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
Here, PD = u, DQ = v,
CP = CQ = d.
In the  PDQ, C is the middle point of PQ
Therefore, (PD)2 + (DQ)2 = 2[R2+d2] where R is the radius
of curvature of the mirror.
Thus, for any path, u2 + v2 = constant = K whatever may be
the position of the point D.
Also,
(u +v)2 = 2(u2 +v2) - (u-v)2
(u+v)2 = 2K- (u-v)2
(i) (u+v)2 is minimum when (u-v)2 is maximum.
When D is at A, (u-v)2 is maximum and is
equal to 2d. When D is at B, (v-u) is equal to 2d
and again (u-v)2 is maximum. Therefore, the
possible paths for minimum, time will be PAQ
and PBQ.
(ii) (u+v)2 is maximum when (u-v)2 = 0. Therefore, u=v. in this
case the point D must lie at O. Thus the path POQ is the
path of maximum time.
Hence, the original statement of Fermat's principle is
modified and it is taken for minimum or maximum time. In such
cases, the principle is known as Femat's principle of stationary
time. Therefore, in its general form the law states that the path
taken by a ray of light in passing from one point to the other is
that of minimum or maximum time.
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
5. THE SIGN CONVENTION
The distance of the object or the image from the refracting
surface is a vector quantity and these distances must be
represented with proper signs. The convention of signs used is
in accordance with the conventions of coordinate geometry as
shown in Fig. 1.5.
Fig. 2.5. Sign Convention
(i) The figures are drawn with the incident light travelling from
left to right.
(ii) The centre of refracting surface is at the origin 0 and its axis
is along XX.'
(iii) Distances measured to the left of 0 are taken as negative.
(iv) Distances measured to the right of 0 are taken as positive.
(v) Distances measured upward and normal to the X-axis are
taken as positive, while downward distances are taken as
negative.
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
In Fig. 1-5, if AB represents an object, its distance OA is
negative. If PQ represents the image, its distance OP is positive.
The size of the object AB is positive while the size of the image
PQ is negative.
Fig. 2.6
6. REFRACTION AT A CONCAVE SURFACE
Let O be a point object in a medium of refractive index 1,
separated from a medium of refractive index 2 by a concave
refracting surface of radius of curvature R. P is the pole of the
surface and OPQ is its axis.
Here, 2 > 1. The image appears to be formed at I.
Considering the area of the image-object OAI (Fig. 2.6),
OAI = OAC - IAC
½ AO.Al sin (i-r) = ½ AO.AC sin i – ½ AL.AC sin r
Taking,
AO = m, AI = n and AC = l
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
½ m.n sin (i-r) = ½ m.l sin i – ½ n.l sin r
mn(sin i cos r-cos i sin r) = ml sin i – nl sin r
1 sin i=2 sin r
From Snell's law,
 mn (
2

sin r cos r-con i sin r ) = ml 2 sin r - nl sin r
1
1
or mn (2 cos r- 1 con i ) = ml 2 - nl 1
Dividing by mnl
 2 cos r   1 cos i
l

2
n

1
... (i)
m
For paraxial rays, m = PO = u,
n = Pl = v, l = PC = R and for small angles;
cos i = cos r = 1
2  1
2

R
v

1
u
Applying the sign convention, u is -ve, v is -ve and R is -ve
2
v

2
v


1
u
1
u


2  1
R
2  1
R
... (ii)
If the first medium is air and the second medium has a
refractive index , then
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES

v

1  1

u
R
(iii)
7. REFRACTION AT A CONVEX SURFACE
(VIRTUAL IMAGE OF A REAL OBJECT)
Here an object is placed in front of a convex surface
separating a medium of refractive index 2 from a medium of
refractive index 1. The virtual image is formed at I (Fig. 2.7).
Considering the area of the image-object OAI,
OAI = IAC - OAC
½ AI.AO sin (i-r) = ½ AI.AC sin r – ½ AO.AC sin i
Taking,
AO = m,
AI = n
and AC = l
½ n.m sin (i-r) = ½ n.l sin r – ½ m.l sin i
nm (sin i cos r-coi i sin r)= nl sin r - ml. sin i
From Snell's law, 1 sin i = 2 sin r
Fig. 2-7.
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Dr M. A. M. EL-Morsy
 nm (
or
REFRACTION THROUGH SPHERICAL SURFACES
2

sin r cos r-con i sin r ) = nl sin r – ml 2 sin r
1
1
nm (2 cos r- 1 con i ) = nl 1 - ml 2
Dividing by mnl
 2 cos r   1 cos i  1

l
m

2
... (i)
n
For paraxial rays, m = PO = u, n = PI = v,
l = PC = R and
for small angles cos i = cos r = 1
1


u
2
v

2  1
R
Applying the sign convention, u is - ve, v is -ve and R is +
ve.
1
u

2
v


2
v
1
u


 2  1
R
2  1
R
... (ii)
If the first medium is air and the second medium in of
refractive index , then 1 = 1 and 2 = 


v

1  1

u
R
8. REFRACTION AT A CONVEX SURFACE
(REAL IMAGE OF A REAL OBJECT)
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
Here, I is the real image of the object O placed in front of a
convex surface separating a medium of refractive index 2 from
a medium of refractive index l.
Fig. 2-8.
Considering the area of the image-object OAI
OAI = IAC + OAC
½ AO.AI sin (i-r) = ½ AI.AC sin r + ½ AO.AC sin i
Taking,
AO = m, AI = n and AC = l
½ m.n sin (i-r) = ½ n.l sin r + ½ m.l sin i
mn (sin i cos r-cos i sin r) = nl sin r + ml. sin i
From Snell's law, 1 sin i = 2 sin r
 mn (
or
2

sin r cos r-con i sin r ) = nl sin r + ml 2 sin r
1
1
mn (2 cos r- 1 con i ) = nl 1 - ml 2
Dividing m.n.l
 2 cos r   1 cos i  1
l

m

2
n
... (i)
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
For paraxial rays, m = PO = u, n = PI = v, l = PC = R and
for small angles cos i = cos r = 1
1


u
2
v

2  1
R
Applying the sign convention, u is - ve, v is +ve and R is +
ve.
1

u
2

v
1

v
2

u

 2  1
R
 2  1
... (ii)
R
If the first medium is air and the second medium in of
refractive index , then 1 = 1 and 2 = 


v

1  1

u
R
Note. When the angles are not small, the correct relation
will be
2
v

1
u

 2 cos r   1 cos i
R
in all the cases.
9. PRINCIPAL FOCI
In the equation
2
infinity then , u =  and
v

1
u

 2  1
R
, if the object is at
1 1
 0 .
u 
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
Let f2 be the value of v corresponding to u = 
2

f2
f2 
2  1
R
2 R
2  1
(i)
This value of v = f2 is termed as the second principal focal
length and the position of the image corresponding to the object
at infinity is called the second principal focus of the refracting
surface.
Similarly, if v =  , i.e. the object is so placed that the
refracted rays become parallel o the principal axis, then
u  f1 
 1R
2  1
(ii)
This value of u is called the first principal focal length of
the refracting surface and is denoted by f1 and the position of the
axial point object, whose image is formed at infinity is known
as the first principal focus of the refracting surface.
From equation (ii) and (i)
f1
1
Adding,
f1
1


R
2  1
f2
2
0
or
,
f2
2

R
2  1
f1 1

f2 2
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
As 1 and 2 are always positive and unequal, f1 and f2 are
always unequal also of oppsite sign.
Further ,
2
v

1
u

 2  1
R
with equations (i) and (ii) we can prove that,
f1 f2

1
u
v
(iii)
This formula holds good only for paraxial rays.
10. Solved Example
Example 1: A concave spherical surface of radius of
curvature 1100 cm separates two media of refractive indices
1.50 and 4/3. An abject is kept in the first medium at a
distance of 30 cm from the surface. Calculate the position of
the image.
Solution
Here 1 = 1.50,
2 = 4/3, R = -100 cm
u = -30 cm, v = ?
2
v

1
u

2  1
R
4 / 3 1.5 4 / 3  1.5


v
30
 100
v = - 27.58 cm
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
The image is formed in the first medium at a distance of
27.58 cm from the refracting surface. The image is virtual.
Example 2. Show that for refraction at a concave spherical
surface (separating glass-air medium), the distance of the object
should be greater than three times the radius of curvature of the
refracting surface for the image to be real.
Solution
Here 1 = 1.50,
2
v

1
u

2 = 1 and R are -ve
2  1
R
Applying the proper signs,
1 1.5 1  1.5


v u
R
,
1 3
1


v 2u 2 R
1 1
3


v 2 R 2u
For v to be positive
1
3

2 R 2u
or
u > 3R
Example 3. A convex surface of radius of curvature 40 cm
separates two media of refractive indices 4/3 and 1.50. An
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
object is kept in the first medium at a distances of 20 cm from
the surface. Calculate the position of the image.
Solution
Here 1 = 4/3,
2 = 1.5, R = + 40 cm
u = -20 cm, v = ?
2
v

1
u

2  1
R
1.5 4 / 3 1.5  ( 4 / 3 )


v
20
40
v = -24 cm
The image is formed in the first medium at a distance of 24
cm from the refracting surface. The image is virtual.
Example 4: A convex refracting surface of radius of curvature
15 cm separates two media of refractive indices 4/3 and 1.50.
An object is kept in the first medium at a distance of 240 cm
from the refracting surface. Calculate the position of the
image.
Solution;
Here 1 = 4/3,
2 = 1.50 , R = + 15 cm
u = -240 cm, v = ?
2
v

1
u

2  1
R
1.5 4 / 3 1.5  ( 4 / 3 )


v
240
15
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
v = 270 cm
The image is formed in the second medium at a distance of
270 cm from the refracting surface. The image is real.
Example 5: The eye can be regarded as a single spherical
refracting surface of radius of curvature of cornea 7.8 mm,
separating two media of refractive indices 1.00 and 1.34.
Calculate the distance from the refracting surface at which a
parallel beam of light will come to focus.
Solution
Here,
1 = 1.00,
2 = 1.34, R = 7.8 mm = 0.78 cm, u
=-
2
v

1
u

2  1
R
1.34 1.00 1.34  1.00


v

0.78
v = 3.075 cm
The image is formed, in the second medium, at a distance of
3.075 cm from the refracting surface. The image is real.
Example 6. A glass dumbell of length 50 cm and refractive
index 1.50 has ends of 5 cm radius of curvature. Find the
position of the image formed due to refraction at one end
only, when a paint object is situated in air, at a distance of
20 cm from the end of the dumbell along the axis.
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REFRACTION THROUGH SPHERICAL SURFACES
Solution;
Fig. 2-9.
2

v
1
u

2  1
R
Here,
1 = 1.00,
2 = 1.50, R = 5 cm, u = - 20, v = ?
1.5 1 1.5  1


v
20
5
or
v = 30 cm
Since v is + ve, the image is formed 30 cm to the right of the
vertex as shown in Fig. 2-9.
Example 7. In Example 6 if the object is 5 cm from the
dumbell, what is the position of the image?
Solution;
Here,
So
u = -5 cm
1.5 1 1.5  1
 
v
5
5
or v = -15 cm
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
Therefore, the image is formed to the left of the vertex as
shown in Fig. (2.10).
Fig. (2-10).
Example 8: A small filament is at the centre of a hollow glass
sphere of inner and outer radii 8 cm and 9 cm respectively.
The refractive index of glass is 1.50, see Fig (2-11).
Calculate the position of the image of the filament when
viewed from outside the sphere.
Fig. (2-11)
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
Solution;
For refraction at the first surface,
u = -8 cm, R1 = - 8 cm , 1 = 1, 2 = 1.5
2
v

1
u

2  1
R1
1.5 1 0.5
 
v 8  8
v’ = -8 cm
It means due to the first surface the image in formed at the
centre.
For the second surface:
u = -9 cm, R2 = - 9 cm , 1 = 1.5, 2 = 1
2
v

1
u

2  1
R2
1 1.5  0.5


v 9
9
v = -9 cm
Thus, the final image is formed at the centre of the sphere.
Example 9: A glass rod has ends as shown in Fig. (2.12). The
refractive index of glass is . The object point O is at a
distance 2R from the surface of larger radius of curvature.
The distance between the apexes of the ends is 3R. Show
that the image point of O is formed at a distance of
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
( 9  4 ) R
( 10   9 )(   2 )
From the right hand vertex.
Fig. (2-12)
Solution;
For the first surface:
2
1

v
u

2  1
R1
Here u = - 2R, R1 = +R , 1 = 1 and

v

v

v 


2 = 
1  1

2R  R
 1
R

1 2  3

2R
2R
2 R
2  3
For the second surface:

 2 R  
 
u    3 R  
2


3



Here R2 = +R/2 , 2 = 1 and
1 = 
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Dr M. A. M. EL-Morsy
REFRACTION THROUGH SPHERICAL SURFACES
2
v

1
u

2  1
R2

1 
1


v 
 2 R   R 2

 
3
R


2


3



 ( 2  3 )
1 2  2


v
R
( 6 R  9 R  2 R )
1 2  2 ( 2  3 )


v
R
R( 4   9 )
1  10  2  29  18

v
R[ 4  9 ]
v
v
( 9  4  )R
10  2  29  18
( 9  4  )R
( 10   9 )(   2 )
Note: In this problem, the image will be real, if the refractive
index of glass is between 2 and 2.25.
The image is virtual, if the refractive index is less than 2 or
it is more than 2.25.
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