Wave Optics - 2 Practice Questions - DPP for NEET

 Page 1


1. (b). The distance of first diffraction minimum from the
central principal maximum x = lD/d
\ sin q = 
x
Dd
l
= Þ d = 
sin
l
q
Þ d = 
8
5000 10
sin30º
-
´
 = 2 × 5 × 10
–5
Þ d = 1.0 × 10
–4
 cm.
2. (b). Here, l
1
 = 5890Å = 5890 × 10
–10
 m
   l
2
 = 5896 Å = 5896 × 10
–10
 m
a =2mm = 2 × 10
–6
 m, D = 2m
For first maxima, sin q = 
11
3x
2aD
l
=
Þ x
1
 = 
1
3D
2a
l
 and x
2
 = 
2
3D
2a
l
\ spacing between the first maxima of two sodium lines
= x
2
 – x
1
 = 
3D
2a
(l
2
 – l
1
)
= 
10
6
3 2(5896 5890) 10
2 2 10
-
-
´ -´
´´
= 9 × 10
–4
 m.
3. (d). 
ax
f
 = nl
l = 
ax
n.f
 = 
33
0.3 10 5 10
31
--
´ ´´
´
l = 5 × 10
–7
 m
l = 5000Å
4. (b). q = sin
–1
 
a
l æö
ç÷
èø
.....(1)
According to question
l = 2 × 10
–3
 m
a = 4 × 10
–3
 m .....(2)
From equation (1) and (2)
q = sin
–1
1
2
æö
ç÷
èø
q = 30°
5. (a). Here the width of principal maxima is 2.5 mm, therefore
its half width is
2.5
22
b
= = 1.25 × 10
–3
m
Diffraction angle q = 
3
/ 2 12.5 10
D2
-
b´
=
\ a q = l
q = l/a = 
3
12.5 10
2
-
´
l = 
3
12.5 10
2
-
´
 × a = 
33
12.5 10 10
2
--
´´
l = 6.25 × 10
–6
 m = 6250 mm
6. (a). Slit width = a = 0.2mm,
b = 
2d
a
l
Angular width W
q
 = 
2
Da
bl
=
q = 
2 6328
0.2
´
= 0.36°
7. (a). Here distance of the screen from the slit,
D = 2m, a = ?, x = 5 mm
    = 5 × 10
–3
 m,l = 5000Å
    = 5000 × 10
–10
 m
For the first minima, sin q = l/a = x/D,
10
3
D 2 5000 10
a
x
5 10
-
-
l ´´
==
´
 = 2 × 10
–4
 m
8. (d). Here, l = 6500Å = 6.5 × 10
–7
 m, a = 0.5 mm =
5 × 10
–4
 m,
D = 1.8 m
Angular separation of two dark bands on each side of central
bright band 2q = 2l/a
Actual distance between them,
2x = 2l/a x D
2x = 
7
4
2 6.5 10 18
5 10
-
-
´ ´´
´
2x = 4.68 × 10
–3
 m
9. (c). Width of central maxima = 
2f
a
l
= 
10
3
2 2 6000 10
0.2 10
-
-
´´´
´
 = 12 mm
10. (a).  q = 
a
l
.....(a)
q = 
x
f
.....(b)
From eqs. (a) and (b)
x
af
l
=
f
x
a
l
= .....(c)
Page 2


1. (b). The distance of first diffraction minimum from the
central principal maximum x = lD/d
\ sin q = 
x
Dd
l
= Þ d = 
sin
l
q
Þ d = 
8
5000 10
sin30º
-
´
 = 2 × 5 × 10
–5
Þ d = 1.0 × 10
–4
 cm.
2. (b). Here, l
1
 = 5890Å = 5890 × 10
–10
 m
   l
2
 = 5896 Å = 5896 × 10
–10
 m
a =2mm = 2 × 10
–6
 m, D = 2m
For first maxima, sin q = 
11
3x
2aD
l
=
Þ x
1
 = 
1
3D
2a
l
 and x
2
 = 
2
3D
2a
l
\ spacing between the first maxima of two sodium lines
= x
2
 – x
1
 = 
3D
2a
(l
2
 – l
1
)
= 
10
6
3 2(5896 5890) 10
2 2 10
-
-
´ -´
´´
= 9 × 10
–4
 m.
3. (d). 
ax
f
 = nl
l = 
ax
n.f
 = 
33
0.3 10 5 10
31
--
´ ´´
´
l = 5 × 10
–7
 m
l = 5000Å
4. (b). q = sin
–1
 
a
l æö
ç÷
èø
.....(1)
According to question
l = 2 × 10
–3
 m
a = 4 × 10
–3
 m .....(2)
From equation (1) and (2)
q = sin
–1
1
2
æö
ç÷
èø
q = 30°
5. (a). Here the width of principal maxima is 2.5 mm, therefore
its half width is
2.5
22
b
= = 1.25 × 10
–3
m
Diffraction angle q = 
3
/ 2 12.5 10
D2
-
b´
=
\ a q = l
q = l/a = 
3
12.5 10
2
-
´
l = 
3
12.5 10
2
-
´
 × a = 
33
12.5 10 10
2
--
´´
l = 6.25 × 10
–6
 m = 6250 mm
6. (a). Slit width = a = 0.2mm,
b = 
2d
a
l
Angular width W
q
 = 
2
Da
bl
=
q = 
2 6328
0.2
´
= 0.36°
7. (a). Here distance of the screen from the slit,
D = 2m, a = ?, x = 5 mm
    = 5 × 10
–3
 m,l = 5000Å
    = 5000 × 10
–10
 m
For the first minima, sin q = l/a = x/D,
10
3
D 2 5000 10
a
x
5 10
-
-
l ´´
==
´
 = 2 × 10
–4
 m
8. (d). Here, l = 6500Å = 6.5 × 10
–7
 m, a = 0.5 mm =
5 × 10
–4
 m,
D = 1.8 m
Angular separation of two dark bands on each side of central
bright band 2q = 2l/a
Actual distance between them,
2x = 2l/a x D
2x = 
7
4
2 6.5 10 18
5 10
-
-
´ ´´
´
2x = 4.68 × 10
–3
 m
9. (c). Width of central maxima = 
2f
a
l
= 
10
3
2 2 6000 10
0.2 10
-
-
´´´
´
 = 12 mm
10. (a).  q = 
a
l
.....(a)
q = 
x
f
.....(b)
From eqs. (a) and (b)
x
af
l
=
f
x
a
l
= .....(c)
DPP/ P 53
149
According to question x = ?, f = 40 cm
l = 5896 × 10
–8
 cm
a = 0.5 × 10
–1
 cm .....(d)
From eqs. (c) and (d)
x = 
8
2
40 5896 10
5 10
-
-
´´
´
 96 = 0.047cm
11. (a). dq = 
1.22
a
l
or a = 
1.22
d
l
q
According to question
dq = 10
–3
 degree = 
3
10
180
-
´p
 Radian,
l = 5 × 10
–5
a = 
5
3
1.22 5 10 180
10 3.14
-
-
´´´
´
a = 3.5 cm
12. (a). Since the reflected light is very highly polarised, it
implies that incident light falls at polarising angle of
incidence q
P
. From Brewster's law ,
m = tanq
p
\q
P
 = tan
–1
 (m) = tan
–1
 (4/3) = 53.1º
Since q
P
 is the angle which the rays from sun make
with the normal to the interface, angle with the interface
will be 90º – 53.1º = 36.9º.
13. (a). Angle of incident light with the surface is 30º. Hence
angle of incidence = 90º – 30º = 60º. Since reflected
light is completely polarised, therefore, incidence takes
place at polarising angle of incidence q
p
.
\  q
p
 = 60º
Using Brewster's law
      m = tan q
p
 = tan 60º
\  m = 
3
14. (d). If unpolarised light is passed through a polariod P
1
, its
intensity will become half.
So I
1
 = 
1
2
 I
0
 with vibrations parallel to the axis of P
1
.
Now this light will pass through second polaroid P
2
whose axis is inclined at ana angle of 30º to the axis of
P
1
 and hence, vibrations of I
1
. So in accordance with
Malus law , the intensity of light emerging from P
2
 will
be
I
2
 = I
1
 cos
2
 30º = 
2
00
1 33
II
2 28
æö
æö
=
ç÷ç÷
èø
èø
2
0
I 3
I8
=
 = 37.5 %
15. (a). If q is the angle between the transmission axes of first
polaroid P
1
 and second P
2
 while f between the
transmission axes of second polaroids P
2
 and P
3
, then
according to given problem,
q + f = 90º or f = (90º – q) ....(1)
Now , if I
0
 is the intensity of unpolarised light incident
on polaroid P
1
, the intensity of light transmitted through
it,
I
1
 =
1
2
I
0
 = I
0
 =
1
2
(32) = 16
2
W
m
....(2)
Now as angle between transmission axes of polaroids
P
1
 and P
2
 is q, in accordance with Malus law , intensity
of light transmitted through P
2
 will be
I
2
 = I
1
 cos
2
q = 16 cos
2
q [from Eq. (2)] ....(3)
And as angle between transmission axes of P
2
 and P
3
is f, light transmitted through P
3
 will be
I
3
 = I
2
cos
2
f = 16 cos
2
q cos
2
f [from Eq.(3)]
Above equation in the light of (1) becomes,
I
3
 = 16 cos
2
q cos
2
 (90º – q) = 4(sin2q)
2
....(4)
According to given problem, I
3
 = 3 W/m
2
So, 4(sin 2q)
2
 = 3   i.e., sin2q = ( 3 /2)
or  2q = 60º i.e. q = 30º
Further in accordance with Eq. (4), I
3
 will be max. when
sin 2q = max., i.e.,
sin 2q = 1 or 2q = 90º, i.e., q = 45º
16. (c) In double refraction light rays always splits into two
rays (O–ray & E–ray). O–ray has same velocity in all
direction but E– ray has different velocity in different
direction.
For calcite m
e 
< m
0 
 Þ v
e
 > v
0
For quartz m
e 
>  m
0 
 Þ v
0
 > v
e
17. (c) At polarizing angle, the reflected and refracted rays
are mutually perpendicular to each other.
18. (d) The amplitude will be A cos60° = A/2
19. (c) Width of central maxima = 
2 D
d
l
= 
7
2
2 2.1 5 10
0.15 10
-
-
´ ´´
´
=1.4×10
–3
 m = 1.4 mm
20. (a) Using d sinq = nl, for n = 1
9
3
550 10
sin
0.55 10
d
-
-
l´
q==
´
=10
–3
 = 0.001 rad
21. (b) A = np dl Þ
A
nd =
pl
= constant 
1
n
d
Þµ
(n = number of blocked HPZ) on decreasing d, n
increases, hence intensity decreases.
22. (b) Intensity of polarized light = 
0
2
I
Þ Intensity of untransmitted light  = 
00
0
22
II
I-=
23. (a)
Page 3


1. (b). The distance of first diffraction minimum from the
central principal maximum x = lD/d
\ sin q = 
x
Dd
l
= Þ d = 
sin
l
q
Þ d = 
8
5000 10
sin30º
-
´
 = 2 × 5 × 10
–5
Þ d = 1.0 × 10
–4
 cm.
2. (b). Here, l
1
 = 5890Å = 5890 × 10
–10
 m
   l
2
 = 5896 Å = 5896 × 10
–10
 m
a =2mm = 2 × 10
–6
 m, D = 2m
For first maxima, sin q = 
11
3x
2aD
l
=
Þ x
1
 = 
1
3D
2a
l
 and x
2
 = 
2
3D
2a
l
\ spacing between the first maxima of two sodium lines
= x
2
 – x
1
 = 
3D
2a
(l
2
 – l
1
)
= 
10
6
3 2(5896 5890) 10
2 2 10
-
-
´ -´
´´
= 9 × 10
–4
 m.
3. (d). 
ax
f
 = nl
l = 
ax
n.f
 = 
33
0.3 10 5 10
31
--
´ ´´
´
l = 5 × 10
–7
 m
l = 5000Å
4. (b). q = sin
–1
 
a
l æö
ç÷
èø
.....(1)
According to question
l = 2 × 10
–3
 m
a = 4 × 10
–3
 m .....(2)
From equation (1) and (2)
q = sin
–1
1
2
æö
ç÷
èø
q = 30°
5. (a). Here the width of principal maxima is 2.5 mm, therefore
its half width is
2.5
22
b
= = 1.25 × 10
–3
m
Diffraction angle q = 
3
/ 2 12.5 10
D2
-
b´
=
\ a q = l
q = l/a = 
3
12.5 10
2
-
´
l = 
3
12.5 10
2
-
´
 × a = 
33
12.5 10 10
2
--
´´
l = 6.25 × 10
–6
 m = 6250 mm
6. (a). Slit width = a = 0.2mm,
b = 
2d
a
l
Angular width W
q
 = 
2
Da
bl
=
q = 
2 6328
0.2
´
= 0.36°
7. (a). Here distance of the screen from the slit,
D = 2m, a = ?, x = 5 mm
    = 5 × 10
–3
 m,l = 5000Å
    = 5000 × 10
–10
 m
For the first minima, sin q = l/a = x/D,
10
3
D 2 5000 10
a
x
5 10
-
-
l ´´
==
´
 = 2 × 10
–4
 m
8. (d). Here, l = 6500Å = 6.5 × 10
–7
 m, a = 0.5 mm =
5 × 10
–4
 m,
D = 1.8 m
Angular separation of two dark bands on each side of central
bright band 2q = 2l/a
Actual distance between them,
2x = 2l/a x D
2x = 
7
4
2 6.5 10 18
5 10
-
-
´ ´´
´
2x = 4.68 × 10
–3
 m
9. (c). Width of central maxima = 
2f
a
l
= 
10
3
2 2 6000 10
0.2 10
-
-
´´´
´
 = 12 mm
10. (a).  q = 
a
l
.....(a)
q = 
x
f
.....(b)
From eqs. (a) and (b)
x
af
l
=
f
x
a
l
= .....(c)
DPP/ P 53
149
According to question x = ?, f = 40 cm
l = 5896 × 10
–8
 cm
a = 0.5 × 10
–1
 cm .....(d)
From eqs. (c) and (d)
x = 
8
2
40 5896 10
5 10
-
-
´´
´
 96 = 0.047cm
11. (a). dq = 
1.22
a
l
or a = 
1.22
d
l
q
According to question
dq = 10
–3
 degree = 
3
10
180
-
´p
 Radian,
l = 5 × 10
–5
a = 
5
3
1.22 5 10 180
10 3.14
-
-
´´´
´
a = 3.5 cm
12. (a). Since the reflected light is very highly polarised, it
implies that incident light falls at polarising angle of
incidence q
P
. From Brewster's law ,
m = tanq
p
\q
P
 = tan
–1
 (m) = tan
–1
 (4/3) = 53.1º
Since q
P
 is the angle which the rays from sun make
with the normal to the interface, angle with the interface
will be 90º – 53.1º = 36.9º.
13. (a). Angle of incident light with the surface is 30º. Hence
angle of incidence = 90º – 30º = 60º. Since reflected
light is completely polarised, therefore, incidence takes
place at polarising angle of incidence q
p
.
\  q
p
 = 60º
Using Brewster's law
      m = tan q
p
 = tan 60º
\  m = 
3
14. (d). If unpolarised light is passed through a polariod P
1
, its
intensity will become half.
So I
1
 = 
1
2
 I
0
 with vibrations parallel to the axis of P
1
.
Now this light will pass through second polaroid P
2
whose axis is inclined at ana angle of 30º to the axis of
P
1
 and hence, vibrations of I
1
. So in accordance with
Malus law , the intensity of light emerging from P
2
 will
be
I
2
 = I
1
 cos
2
 30º = 
2
00
1 33
II
2 28
æö
æö
=
ç÷ç÷
èø
èø
2
0
I 3
I8
=
 = 37.5 %
15. (a). If q is the angle between the transmission axes of first
polaroid P
1
 and second P
2
 while f between the
transmission axes of second polaroids P
2
 and P
3
, then
according to given problem,
q + f = 90º or f = (90º – q) ....(1)
Now , if I
0
 is the intensity of unpolarised light incident
on polaroid P
1
, the intensity of light transmitted through
it,
I
1
 =
1
2
I
0
 = I
0
 =
1
2
(32) = 16
2
W
m
....(2)
Now as angle between transmission axes of polaroids
P
1
 and P
2
 is q, in accordance with Malus law , intensity
of light transmitted through P
2
 will be
I
2
 = I
1
 cos
2
q = 16 cos
2
q [from Eq. (2)] ....(3)
And as angle between transmission axes of P
2
 and P
3
is f, light transmitted through P
3
 will be
I
3
 = I
2
cos
2
f = 16 cos
2
q cos
2
f [from Eq.(3)]
Above equation in the light of (1) becomes,
I
3
 = 16 cos
2
q cos
2
 (90º – q) = 4(sin2q)
2
....(4)
According to given problem, I
3
 = 3 W/m
2
So, 4(sin 2q)
2
 = 3   i.e., sin2q = ( 3 /2)
or  2q = 60º i.e. q = 30º
Further in accordance with Eq. (4), I
3
 will be max. when
sin 2q = max., i.e.,
sin 2q = 1 or 2q = 90º, i.e., q = 45º
16. (c) In double refraction light rays always splits into two
rays (O–ray & E–ray). O–ray has same velocity in all
direction but E– ray has different velocity in different
direction.
For calcite m
e 
< m
0 
 Þ v
e
 > v
0
For quartz m
e 
>  m
0 
 Þ v
0
 > v
e
17. (c) At polarizing angle, the reflected and refracted rays
are mutually perpendicular to each other.
18. (d) The amplitude will be A cos60° = A/2
19. (c) Width of central maxima = 
2 D
d
l
= 
7
2
2 2.1 5 10
0.15 10
-
-
´ ´´
´
=1.4×10
–3
 m = 1.4 mm
20. (a) Using d sinq = nl, for n = 1
9
3
550 10
sin
0.55 10
d
-
-
l´
q==
´
=10
–3
 = 0.001 rad
21. (b) A = np dl Þ
A
nd =
pl
= constant 
1
n
d
Þµ
(n = number of blocked HPZ) on decreasing d, n
increases, hence intensity decreases.
22. (b) Intensity of polarized light = 
0
2
I
Þ Intensity of untransmitted light  = 
00
0
22
II
I-=
23. (a)
DPP/ P 53
150
24. (a) It magnitude of light vector varies periodically during
it's rotation, the tip of vector traces an ellipse and light
is said to be elliptically polarised. This is not in nicol
prism.
25. (a) Multiple focii of zone plate given by 
( )
2
21
n
p
r
f
p
=
-l
,
where p = 1, 2, 3 ........
26. (a) Angular width is the angle subtended by the distance
between first minima on either side at the centre of the
slit. It is given by f = 2 q , where q is the angle of
diffraction.
For first diffraction minimum,   a sin q = 1 l
or     sin q = l/a     or    
a
l
= q
\ Angular width   
2
2 i.e.
a
l
f= q= fµl
2
1
2
1
l
l
=
f
f
;  Å 4200
100
70
6000
1
2
1 2
= ´ =
f
f
l = l \
27. (b) On immersing in liquid, a wavelength l = 6000 Å must
be behaving as l' = 4200 Å to get the same decrease in
angular width. Therefore, refractive index of medium
. 43 . 1
4200
6000
= =
l¢
l
= m
28. (a) When a polaroid is rotated in the path of unpolarised
light, the intensity of light transmitted from polaroid
remains undiminished (because unpolarised light
contains waves vibrating in all possible planes with
rotated in path of plane polarised light, its intensity
will vary from maximum (when the vibrations of the
plane polarised light are parallel to the axis of the
polaroid) to minimum (when the direction of the
vibrations becomes perpendicular to the axis of the
crystal). Thus using polaroid we can easily verify that
whether the light is polarised or not.
29. (d) The nicol prism is made of calcite crystal. When light
is  passed through calcite crystal, it breaks up into two
rays
(i) the ordinary ray which has its electric vector
perpendicular to the principal section of the crystal and
(ii) the extra ordinary ray which has its electric vector
parallel to the principal section. The nicol prism is made
in such a way that it eliminates one of the two rays by
total internal reflection, thus produces plane polarised
light. It is generally found that the ordinary ray is
eliminated and only the extra ordinary ray is transmitted
through the prism. The nicol prism consists of two
calcite crystal cut at - 68° with its principal axis joined
by a glue called Canada balsam.
30. (d) The clouds consists of dust particles and water droplets.
Their size is very large as compared to the wavelength
of the incident light from the sun. So there is very little
scattering of light. Hence the light which we receive
through the clouds has all the colours of light. As a
result of this, we receive almost white light. Therefore,
the cloud are generally white.