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9 | 139-142 | 8)
h and h¢ will be taken positive or negative in accordance with the accepted
sign convention In triangles A¢B¢P and ABP, we have,
B A
B P
BA
BP
′
′
′
=
With the sign convention, this becomes
Rationalised 2023-24
Physics
226
–
–
h
v
h
u
′ = –
so that
m =
–
h
v
h
u
′ =
(9 9)
We have derived here the mirror equation, Eq (9 |
9 | 140-143 | In triangles A¢B¢P and ABP, we have,
B A
B P
BA
BP
′
′
′
=
With the sign convention, this becomes
Rationalised 2023-24
Physics
226
–
–
h
v
h
u
′ = –
so that
m =
–
h
v
h
u
′ =
(9 9)
We have derived here the mirror equation, Eq (9 7), and the
magnification formula, Eq |
9 | 141-144 | 9)
We have derived here the mirror equation, Eq (9 7), and the
magnification formula, Eq (9 |
9 | 142-145 | (9 7), and the
magnification formula, Eq (9 9), for the case of real, inverted image formed
by a concave mirror |
9 | 143-146 | 7), and the
magnification formula, Eq (9 9), for the case of real, inverted image formed
by a concave mirror With the proper use of sign convention, these are,
in fact, valid for all the cases of reflection by a spherical mirror (concave
or convex) whether the image formed is real or virtual |
9 | 144-147 | (9 9), for the case of real, inverted image formed
by a concave mirror With the proper use of sign convention, these are,
in fact, valid for all the cases of reflection by a spherical mirror (concave
or convex) whether the image formed is real or virtual Figure 9 |
9 | 145-148 | 9), for the case of real, inverted image formed
by a concave mirror With the proper use of sign convention, these are,
in fact, valid for all the cases of reflection by a spherical mirror (concave
or convex) whether the image formed is real or virtual Figure 9 6 shows
the ray diagrams for virtual image formed by a concave and convex mirror |
9 | 146-149 | With the proper use of sign convention, these are,
in fact, valid for all the cases of reflection by a spherical mirror (concave
or convex) whether the image formed is real or virtual Figure 9 6 shows
the ray diagrams for virtual image formed by a concave and convex mirror You should verify that Eqs |
9 | 147-150 | Figure 9 6 shows
the ray diagrams for virtual image formed by a concave and convex mirror You should verify that Eqs (9 |
9 | 148-151 | 6 shows
the ray diagrams for virtual image formed by a concave and convex mirror You should verify that Eqs (9 7) and (9 |
9 | 149-152 | You should verify that Eqs (9 7) and (9 9) are valid for these cases as
well |
9 | 150-153 | (9 7) and (9 9) are valid for these cases as
well FIGURE 9 |
9 | 151-154 | 7) and (9 9) are valid for these cases as
well FIGURE 9 6 Image formation by (a) a concave mirror with object between
P and F, and (b) a convex mirror |
9 | 152-155 | 9) are valid for these cases as
well FIGURE 9 6 Image formation by (a) a concave mirror with object between
P and F, and (b) a convex mirror EXAMPLE 9 |
9 | 153-156 | FIGURE 9 6 Image formation by (a) a concave mirror with object between
P and F, and (b) a convex mirror EXAMPLE 9 1
Example 9 |
9 | 154-157 | 6 Image formation by (a) a concave mirror with object between
P and F, and (b) a convex mirror EXAMPLE 9 1
Example 9 1 Suppose that the lower half of the concave mirror’s
reflecting surface in Fig |
9 | 155-158 | EXAMPLE 9 1
Example 9 1 Suppose that the lower half of the concave mirror’s
reflecting surface in Fig 9 |
9 | 156-159 | 1
Example 9 1 Suppose that the lower half of the concave mirror’s
reflecting surface in Fig 9 6 is covered with an opaque (non-reflective)
material |
9 | 157-160 | 1 Suppose that the lower half of the concave mirror’s
reflecting surface in Fig 9 6 is covered with an opaque (non-reflective)
material What effect will this have on the image of an object placed
in front of the mirror |
9 | 158-161 | 9 6 is covered with an opaque (non-reflective)
material What effect will this have on the image of an object placed
in front of the mirror Solution You may think that the image will now show only half of the
object, but taking the laws of reflection to be true for all points of the
remaining part of the mirror, the image will be that of the whole object |
9 | 159-162 | 6 is covered with an opaque (non-reflective)
material What effect will this have on the image of an object placed
in front of the mirror Solution You may think that the image will now show only half of the
object, but taking the laws of reflection to be true for all points of the
remaining part of the mirror, the image will be that of the whole object However, as the area of the reflecting surface has been reduced, the
intensity of the image will be low (in this case, half) |
9 | 160-163 | What effect will this have on the image of an object placed
in front of the mirror Solution You may think that the image will now show only half of the
object, but taking the laws of reflection to be true for all points of the
remaining part of the mirror, the image will be that of the whole object However, as the area of the reflecting surface has been reduced, the
intensity of the image will be low (in this case, half) Example 9 |
9 | 161-164 | Solution You may think that the image will now show only half of the
object, but taking the laws of reflection to be true for all points of the
remaining part of the mirror, the image will be that of the whole object However, as the area of the reflecting surface has been reduced, the
intensity of the image will be low (in this case, half) Example 9 2 A mobile phone lies along the principal axis of a concave
mirror, as shown in Fig |
9 | 162-165 | However, as the area of the reflecting surface has been reduced, the
intensity of the image will be low (in this case, half) Example 9 2 A mobile phone lies along the principal axis of a concave
mirror, as shown in Fig 9 |
9 | 163-166 | Example 9 2 A mobile phone lies along the principal axis of a concave
mirror, as shown in Fig 9 7 |
9 | 164-167 | 2 A mobile phone lies along the principal axis of a concave
mirror, as shown in Fig 9 7 Show by suitable diagram, the formation
of its image |
9 | 165-168 | 9 7 Show by suitable diagram, the formation
of its image Explain why the magnification is not uniform |
9 | 166-169 | 7 Show by suitable diagram, the formation
of its image Explain why the magnification is not uniform Will the
distortion of image depend on the location of the phone with respect
to the mirror |
9 | 167-170 | Show by suitable diagram, the formation
of its image Explain why the magnification is not uniform Will the
distortion of image depend on the location of the phone with respect
to the mirror FIGURE 9 |
9 | 168-171 | Explain why the magnification is not uniform Will the
distortion of image depend on the location of the phone with respect
to the mirror FIGURE 9 7
EXAMPLE 9 |
9 | 169-172 | Will the
distortion of image depend on the location of the phone with respect
to the mirror FIGURE 9 7
EXAMPLE 9 2
Rationalised 2023-24
Ray Optics and
Optical Instruments
227
EXAMPLE 9 |
9 | 170-173 | FIGURE 9 7
EXAMPLE 9 2
Rationalised 2023-24
Ray Optics and
Optical Instruments
227
EXAMPLE 9 3
EXAMPLE 9 |
9 | 171-174 | 7
EXAMPLE 9 2
Rationalised 2023-24
Ray Optics and
Optical Instruments
227
EXAMPLE 9 3
EXAMPLE 9 4
Solution
The ray diagram for the formation of the image of the phone is shown
in Fig |
9 | 172-175 | 2
Rationalised 2023-24
Ray Optics and
Optical Instruments
227
EXAMPLE 9 3
EXAMPLE 9 4
Solution
The ray diagram for the formation of the image of the phone is shown
in Fig 9 |
9 | 173-176 | 3
EXAMPLE 9 4
Solution
The ray diagram for the formation of the image of the phone is shown
in Fig 9 7 |
9 | 174-177 | 4
Solution
The ray diagram for the formation of the image of the phone is shown
in Fig 9 7 The image of the part which is on the plane perpendicular
to principal axis will be on the same plane |
9 | 175-178 | 9 7 The image of the part which is on the plane perpendicular
to principal axis will be on the same plane It will be of the same size,
i |
9 | 176-179 | 7 The image of the part which is on the plane perpendicular
to principal axis will be on the same plane It will be of the same size,
i e |
9 | 177-180 | The image of the part which is on the plane perpendicular
to principal axis will be on the same plane It will be of the same size,
i e , B¢C = BC |
9 | 178-181 | It will be of the same size,
i e , B¢C = BC You can yourself realise why the image is distorted |
9 | 179-182 | e , B¢C = BC You can yourself realise why the image is distorted Example 9 |
9 | 180-183 | , B¢C = BC You can yourself realise why the image is distorted Example 9 3
An object is placed at (i) 10 cm, (ii) 5 cm in front of a
concave mirror of radius of curvature 15 cm |
9 | 181-184 | You can yourself realise why the image is distorted Example 9 3
An object is placed at (i) 10 cm, (ii) 5 cm in front of a
concave mirror of radius of curvature 15 cm Find the position, nature,
and magnification of the image in each case |
9 | 182-185 | Example 9 3
An object is placed at (i) 10 cm, (ii) 5 cm in front of a
concave mirror of radius of curvature 15 cm Find the position, nature,
and magnification of the image in each case Solution
The focal length f = –15/2 cm = –7 |
9 | 183-186 | 3
An object is placed at (i) 10 cm, (ii) 5 cm in front of a
concave mirror of radius of curvature 15 cm Find the position, nature,
and magnification of the image in each case Solution
The focal length f = –15/2 cm = –7 5 cm
(i) The object distance u = –10 cm |
9 | 184-187 | Find the position, nature,
and magnification of the image in each case Solution
The focal length f = –15/2 cm = –7 5 cm
(i) The object distance u = –10 cm Then Eq |
9 | 185-188 | Solution
The focal length f = –15/2 cm = –7 5 cm
(i) The object distance u = –10 cm Then Eq (9 |
9 | 186-189 | 5 cm
(i) The object distance u = –10 cm Then Eq (9 7) gives
–
– |
9 | 187-190 | Then Eq (9 7) gives
–
– 1
1
1
10
7 5
v +
=
or |
9 | 188-191 | (9 7) gives
–
– 1
1
1
10
7 5
v +
=
or 10
7 5
2 5
v
−×
=
= – 30 cm
The image is 30 cm from the mirror on the same side as the object |
9 | 189-192 | 7) gives
–
– 1
1
1
10
7 5
v +
=
or 10
7 5
2 5
v
−×
=
= – 30 cm
The image is 30 cm from the mirror on the same side as the object Also, magnification m =
( 30)
–
–
– 3
( 10)
v
u
−
=
=
−
The image is magnified, real and inverted |
9 | 190-193 | 1
1
1
10
7 5
v +
=
or 10
7 5
2 5
v
−×
=
= – 30 cm
The image is 30 cm from the mirror on the same side as the object Also, magnification m =
( 30)
–
–
– 3
( 10)
v
u
−
=
=
−
The image is magnified, real and inverted (ii) The object distance u = –5 cm |
9 | 191-194 | 10
7 5
2 5
v
−×
=
= – 30 cm
The image is 30 cm from the mirror on the same side as the object Also, magnification m =
( 30)
–
–
– 3
( 10)
v
u
−
=
=
−
The image is magnified, real and inverted (ii) The object distance u = –5 cm Then from Eq |
9 | 192-195 | Also, magnification m =
( 30)
–
–
– 3
( 10)
v
u
−
=
=
−
The image is magnified, real and inverted (ii) The object distance u = –5 cm Then from Eq (9 |
9 | 193-196 | (ii) The object distance u = –5 cm Then from Eq (9 7),
1
1
1
5
7 |
9 | 194-197 | Then from Eq (9 7),
1
1
1
5
7 5
v +
=
−
−
or
(
) |
9 | 195-198 | (9 7),
1
1
1
5
7 5
v +
=
−
−
or
(
) –
5
7 5
15 cm
7 5
5
v
×
=
=
This image is formed at 15 cm behind the mirror |
9 | 196-199 | 7),
1
1
1
5
7 5
v +
=
−
−
or
(
) –
5
7 5
15 cm
7 5
5
v
×
=
=
This image is formed at 15 cm behind the mirror It is a virtual image |
9 | 197-200 | 5
v +
=
−
−
or
(
) –
5
7 5
15 cm
7 5
5
v
×
=
=
This image is formed at 15 cm behind the mirror It is a virtual image Magnification m =
15
–
–
3
( 5)
v
u =
=
−
The image is magnified, virtual and erect |
9 | 198-201 | –
5
7 5
15 cm
7 5
5
v
×
=
=
This image is formed at 15 cm behind the mirror It is a virtual image Magnification m =
15
–
–
3
( 5)
v
u =
=
−
The image is magnified, virtual and erect Example 9 |
9 | 199-202 | It is a virtual image Magnification m =
15
–
–
3
( 5)
v
u =
=
−
The image is magnified, virtual and erect Example 9 4 Suppose while sitting in a parked car, you notice a
jogger approaching towards you in the side view mirror of R = 2 m |
9 | 200-203 | Magnification m =
15
–
–
3
( 5)
v
u =
=
−
The image is magnified, virtual and erect Example 9 4 Suppose while sitting in a parked car, you notice a
jogger approaching towards you in the side view mirror of R = 2 m If
the jogger is running at a speed of 5 m s–1, how fast the image of the
jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m,
and (d) 9 m away |
9 | 201-204 | Example 9 4 Suppose while sitting in a parked car, you notice a
jogger approaching towards you in the side view mirror of R = 2 m If
the jogger is running at a speed of 5 m s–1, how fast the image of the
jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m,
and (d) 9 m away Solution
From the mirror equation, Eq |
9 | 202-205 | 4 Suppose while sitting in a parked car, you notice a
jogger approaching towards you in the side view mirror of R = 2 m If
the jogger is running at a speed of 5 m s–1, how fast the image of the
jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m,
and (d) 9 m away Solution
From the mirror equation, Eq (9 |
9 | 203-206 | If
the jogger is running at a speed of 5 m s–1, how fast the image of the
jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m,
and (d) 9 m away Solution
From the mirror equation, Eq (9 7), we get
fu
v
u
f
=
−
For convex mirror, since R = 2 m, f = 1 m |
9 | 204-207 | Solution
From the mirror equation, Eq (9 7), we get
fu
v
u
f
=
−
For convex mirror, since R = 2 m, f = 1 m Then
for u = –39 m,
( 39)
1
39 m
39
1
40
v
−
×
=
=
−
−
Since the jogger moves at a constant speed of 5 m s–1, after 1 s the
position of the image v (for u = –39 + 5 = –34) is (34/35 )m |
9 | 205-208 | (9 7), we get
fu
v
u
f
=
−
For convex mirror, since R = 2 m, f = 1 m Then
for u = –39 m,
( 39)
1
39 m
39
1
40
v
−
×
=
=
−
−
Since the jogger moves at a constant speed of 5 m s–1, after 1 s the
position of the image v (for u = –39 + 5 = –34) is (34/35 )m EXAMPLE 9 |
9 | 206-209 | 7), we get
fu
v
u
f
=
−
For convex mirror, since R = 2 m, f = 1 m Then
for u = –39 m,
( 39)
1
39 m
39
1
40
v
−
×
=
=
−
−
Since the jogger moves at a constant speed of 5 m s–1, after 1 s the
position of the image v (for u = –39 + 5 = –34) is (34/35 )m EXAMPLE 9 2
Rationalised 2023-24
Physics
228
EXAMPLE 9 |
9 | 207-210 | Then
for u = –39 m,
( 39)
1
39 m
39
1
40
v
−
×
=
=
−
−
Since the jogger moves at a constant speed of 5 m s–1, after 1 s the
position of the image v (for u = –39 + 5 = –34) is (34/35 )m EXAMPLE 9 2
Rationalised 2023-24
Physics
228
EXAMPLE 9 4
FIGURE 9 |
9 | 208-211 | EXAMPLE 9 2
Rationalised 2023-24
Physics
228
EXAMPLE 9 4
FIGURE 9 8 Refraction and reflection of light |
9 | 209-212 | 2
Rationalised 2023-24
Physics
228
EXAMPLE 9 4
FIGURE 9 8 Refraction and reflection of light The shift in the position of image in 1 s is
1365
1360
39
34
5
1
m
40
35
1400
1400
280
−
−
=
=
=
Therefore, the average speed of the image when the jogger is between
39 m and 34 m from the mirror, is (1/280) m s–1
Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the
speed with which the image appears to move is
–1
–1
–1
1
1
1
m s ,
m s
and
m s ,
150
60
10
respectively |
9 | 210-213 | 4
FIGURE 9 8 Refraction and reflection of light The shift in the position of image in 1 s is
1365
1360
39
34
5
1
m
40
35
1400
1400
280
−
−
=
=
=
Therefore, the average speed of the image when the jogger is between
39 m and 34 m from the mirror, is (1/280) m s–1
Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the
speed with which the image appears to move is
–1
–1
–1
1
1
1
m s ,
m s
and
m s ,
150
60
10
respectively Although the jogger has been moving with a constant speed, the speed
of his/her image appears to increase substantially as he/she moves
closer to the mirror |
9 | 211-214 | 8 Refraction and reflection of light The shift in the position of image in 1 s is
1365
1360
39
34
5
1
m
40
35
1400
1400
280
−
−
=
=
=
Therefore, the average speed of the image when the jogger is between
39 m and 34 m from the mirror, is (1/280) m s–1
Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the
speed with which the image appears to move is
–1
–1
–1
1
1
1
m s ,
m s
and
m s ,
150
60
10
respectively Although the jogger has been moving with a constant speed, the speed
of his/her image appears to increase substantially as he/she moves
closer to the mirror This phenomenon can be noticed by any person
sitting in a stationary car or a bus |
9 | 212-215 | The shift in the position of image in 1 s is
1365
1360
39
34
5
1
m
40
35
1400
1400
280
−
−
=
=
=
Therefore, the average speed of the image when the jogger is between
39 m and 34 m from the mirror, is (1/280) m s–1
Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the
speed with which the image appears to move is
–1
–1
–1
1
1
1
m s ,
m s
and
m s ,
150
60
10
respectively Although the jogger has been moving with a constant speed, the speed
of his/her image appears to increase substantially as he/she moves
closer to the mirror This phenomenon can be noticed by any person
sitting in a stationary car or a bus In case of moving vehicles, a
similar phenomenon could be observed if the vehicle in the rear is
moving closer with a constant speed |
9 | 213-216 | Although the jogger has been moving with a constant speed, the speed
of his/her image appears to increase substantially as he/she moves
closer to the mirror This phenomenon can be noticed by any person
sitting in a stationary car or a bus In case of moving vehicles, a
similar phenomenon could be observed if the vehicle in the rear is
moving closer with a constant speed 9 |
9 | 214-217 | This phenomenon can be noticed by any person
sitting in a stationary car or a bus In case of moving vehicles, a
similar phenomenon could be observed if the vehicle in the rear is
moving closer with a constant speed 9 3 REFRACTION
When a beam of light encounters another transparent medium, a part of
light gets reflected back into the first medium while the rest enters the
other |
9 | 215-218 | In case of moving vehicles, a
similar phenomenon could be observed if the vehicle in the rear is
moving closer with a constant speed 9 3 REFRACTION
When a beam of light encounters another transparent medium, a part of
light gets reflected back into the first medium while the rest enters the
other A ray of light represents a beam |
9 | 216-219 | 9 3 REFRACTION
When a beam of light encounters another transparent medium, a part of
light gets reflected back into the first medium while the rest enters the
other A ray of light represents a beam The direction of propagation of an
obliquely incident (0°< i < 90°) ray of light that enters the other medium,
changes at the interface of the two media |
9 | 217-220 | 3 REFRACTION
When a beam of light encounters another transparent medium, a part of
light gets reflected back into the first medium while the rest enters the
other A ray of light represents a beam The direction of propagation of an
obliquely incident (0°< i < 90°) ray of light that enters the other medium,
changes at the interface of the two media This phenomenon is called
refraction of light |
9 | 218-221 | A ray of light represents a beam The direction of propagation of an
obliquely incident (0°< i < 90°) ray of light that enters the other medium,
changes at the interface of the two media This phenomenon is called
refraction of light Snell experimentally obtained the following laws
of refraction:
(i)
The incident ray, the refracted ray and the
normal to the interface at the point of
incidence, all lie in the same plane |
9 | 219-222 | The direction of propagation of an
obliquely incident (0°< i < 90°) ray of light that enters the other medium,
changes at the interface of the two media This phenomenon is called
refraction of light Snell experimentally obtained the following laws
of refraction:
(i)
The incident ray, the refracted ray and the
normal to the interface at the point of
incidence, all lie in the same plane (ii) The ratio of the sine of the angle of incidence
to the sine of angle of refraction is constant |
9 | 220-223 | This phenomenon is called
refraction of light Snell experimentally obtained the following laws
of refraction:
(i)
The incident ray, the refracted ray and the
normal to the interface at the point of
incidence, all lie in the same plane (ii) The ratio of the sine of the angle of incidence
to the sine of angle of refraction is constant Remember that the angles of incidence (i ) and
refraction (r ) are the angles that the incident
and its refracted ray make with the normal,
respectively |
9 | 221-224 | Snell experimentally obtained the following laws
of refraction:
(i)
The incident ray, the refracted ray and the
normal to the interface at the point of
incidence, all lie in the same plane (ii) The ratio of the sine of the angle of incidence
to the sine of angle of refraction is constant Remember that the angles of incidence (i ) and
refraction (r ) are the angles that the incident
and its refracted ray make with the normal,
respectively We have
21
sin
sin
i
r =n
(9 |
9 | 222-225 | (ii) The ratio of the sine of the angle of incidence
to the sine of angle of refraction is constant Remember that the angles of incidence (i ) and
refraction (r ) are the angles that the incident
and its refracted ray make with the normal,
respectively We have
21
sin
sin
i
r =n
(9 10)
where n 21 is a constant, called the refractive
index of the second medium with respect to the
first medium |
9 | 223-226 | Remember that the angles of incidence (i ) and
refraction (r ) are the angles that the incident
and its refracted ray make with the normal,
respectively We have
21
sin
sin
i
r =n
(9 10)
where n 21 is a constant, called the refractive
index of the second medium with respect to the
first medium Equation (9 |
9 | 224-227 | We have
21
sin
sin
i
r =n
(9 10)
where n 21 is a constant, called the refractive
index of the second medium with respect to the
first medium Equation (9 10) is the well-known
Snell’s law of refraction |
9 | 225-228 | 10)
where n 21 is a constant, called the refractive
index of the second medium with respect to the
first medium Equation (9 10) is the well-known
Snell’s law of refraction We note that n 21 is a
characteristic of the pair of media (and also depends on the wavelength
of light), but is independent of the angle of incidence |
9 | 226-229 | Equation (9 10) is the well-known
Snell’s law of refraction We note that n 21 is a
characteristic of the pair of media (and also depends on the wavelength
of light), but is independent of the angle of incidence From Eq |
9 | 227-230 | 10) is the well-known
Snell’s law of refraction We note that n 21 is a
characteristic of the pair of media (and also depends on the wavelength
of light), but is independent of the angle of incidence From Eq (9 |
9 | 228-231 | We note that n 21 is a
characteristic of the pair of media (and also depends on the wavelength
of light), but is independent of the angle of incidence From Eq (9 10), if n 21 > 1, r < i, i |
9 | 229-232 | From Eq (9 10), if n 21 > 1, r < i, i e |
9 | 230-233 | (9 10), if n 21 > 1, r < i, i e , the refracted ray bends towards
the normal |
9 | 231-234 | 10), if n 21 > 1, r < i, i e , the refracted ray bends towards
the normal In such a case medium 2 is said to be optically denser (or
denser, in short) than medium 1 |
9 | 232-235 | e , the refracted ray bends towards
the normal In such a case medium 2 is said to be optically denser (or
denser, in short) than medium 1 On the other hand, if n 21 <1, r > i, the
Rationalised 2023-24
Ray Optics and
Optical Instruments
229
refracted ray bends away from the normal |
9 | 233-236 | , the refracted ray bends towards
the normal In such a case medium 2 is said to be optically denser (or
denser, in short) than medium 1 On the other hand, if n 21 <1, r > i, the
Rationalised 2023-24
Ray Optics and
Optical Instruments
229
refracted ray bends away from the normal This
is the case when incident ray in a denser
medium refracts into a rarer medium |
9 | 234-237 | In such a case medium 2 is said to be optically denser (or
denser, in short) than medium 1 On the other hand, if n 21 <1, r > i, the
Rationalised 2023-24
Ray Optics and
Optical Instruments
229
refracted ray bends away from the normal This
is the case when incident ray in a denser
medium refracts into a rarer medium Note: Optical density should not be
confused with mass density, which is mass
per unit volume |
9 | 235-238 | On the other hand, if n 21 <1, r > i, the
Rationalised 2023-24
Ray Optics and
Optical Instruments
229
refracted ray bends away from the normal This
is the case when incident ray in a denser
medium refracts into a rarer medium Note: Optical density should not be
confused with mass density, which is mass
per unit volume It is possible that mass
density of an optically denser medium may
be less than that of an optically rarer
medium (optical density is the ratio of the
speed of light in two media) |
9 | 236-239 | This
is the case when incident ray in a denser
medium refracts into a rarer medium Note: Optical density should not be
confused with mass density, which is mass
per unit volume It is possible that mass
density of an optically denser medium may
be less than that of an optically rarer
medium (optical density is the ratio of the
speed of light in two media) For example,
turpentine and water |
9 | 237-240 | Note: Optical density should not be
confused with mass density, which is mass
per unit volume It is possible that mass
density of an optically denser medium may
be less than that of an optically rarer
medium (optical density is the ratio of the
speed of light in two media) For example,
turpentine and water Mass density of
turpentine is less than that of water but
its optical density is higher |
9 | 238-241 | It is possible that mass
density of an optically denser medium may
be less than that of an optically rarer
medium (optical density is the ratio of the
speed of light in two media) For example,
turpentine and water Mass density of
turpentine is less than that of water but
its optical density is higher If n 21 is the refractive index of medium 2 with
respect to medium 1 and n12 the refractive index
of medium 1 with respect to medium 2, then it
should be clear that
12
21
1
n
=n
(9 |
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