Wave Optics Practice Questions - DPP for JEE

 Page 1


1. (d) Optical path difference
?x = .
2. (b)
3. (a) For a circularly polarised light electric field remains constant with
time.
4. (b) Let nth fringe of 2500 Å coincide with (n – 2)th fringe of 3500Å.
?  3500 (n – 2) = 2500 × n
1000 n = 7000, n = 7
?  7th order fringe of 1st source will coincide with 5th order
fringe of 2nd source.
5. (d) Let P is a point infront of one slit at which intensity is to be
calculated. From figure,
Path difference = S
2
P – S
1
P
=  = 
= 
?x = 
Page 2


1. (d) Optical path difference
?x = .
2. (b)
3. (a) For a circularly polarised light electric field remains constant with
time.
4. (b) Let nth fringe of 2500 Å coincide with (n – 2)th fringe of 3500Å.
?  3500 (n – 2) = 2500 × n
1000 n = 7000, n = 7
?  7th order fringe of 1st source will coincide with 5th order
fringe of 2nd source.
5. (d) Let P is a point infront of one slit at which intensity is to be
calculated. From figure,
Path difference = S
2
P – S
1
P
=  = 
= 
?x = 
Place difference,
?f = 
So, resultant intensity at the desired point ‘p’ is
I = I
0
cos
2
 = I
0
cos
2
6. (b) = 
and = .
As > , so shift will be along –ve y-axis.
7. (b) The width of the central maximum is given by
If  then ß decreases.
Also, intensity 
where                
I increases as d increases
The central maximum will become narrower and brighter.
8. (c) Here Angle of incidence, i = 57
tan 57° = 1.54
u
glass
 = tan i
It means, Here Brewster’s law is followed and the reflected ray is
completely polarised.
Now, when reflected ray is analysed through a polaroid then intensity of
light is given by malus law.
i.e. I = I
0
 cos
2
?
on rotating polaroid ‘?’ changes. Due to which intensity first decreases
and then increases.
9. (c) Here 
Page 3


1. (d) Optical path difference
?x = .
2. (b)
3. (a) For a circularly polarised light electric field remains constant with
time.
4. (b) Let nth fringe of 2500 Å coincide with (n – 2)th fringe of 3500Å.
?  3500 (n – 2) = 2500 × n
1000 n = 7000, n = 7
?  7th order fringe of 1st source will coincide with 5th order
fringe of 2nd source.
5. (d) Let P is a point infront of one slit at which intensity is to be
calculated. From figure,
Path difference = S
2
P – S
1
P
=  = 
= 
?x = 
Place difference,
?f = 
So, resultant intensity at the desired point ‘p’ is
I = I
0
cos
2
 = I
0
cos
2
6. (b) = 
and = .
As > , so shift will be along –ve y-axis.
7. (b) The width of the central maximum is given by
If  then ß decreases.
Also, intensity 
where                
I increases as d increases
The central maximum will become narrower and brighter.
8. (c) Here Angle of incidence, i = 57
tan 57° = 1.54
u
glass
 = tan i
It means, Here Brewster’s law is followed and the reflected ray is
completely polarised.
Now, when reflected ray is analysed through a polaroid then intensity of
light is given by malus law.
i.e. I = I
0
 cos
2
?
on rotating polaroid ‘?’ changes. Due to which intensity first decreases
and then increases.
9. (c) Here 
Q a
1
 = a
2
 = a
? 
Now,      ? 
.
10. (c) For path difference ?, phase difference= 2p rad.
For path difference , phase difference =  rad.
As K = 4I
0
  so intensity at given point where path difference is 
K' = 
= 
11. (c)
12. (c)
13. (b) For first minima at P
AP – BP = ?
AP – MP = 
Page 4


1. (d) Optical path difference
?x = .
2. (b)
3. (a) For a circularly polarised light electric field remains constant with
time.
4. (b) Let nth fringe of 2500 Å coincide with (n – 2)th fringe of 3500Å.
?  3500 (n – 2) = 2500 × n
1000 n = 7000, n = 7
?  7th order fringe of 1st source will coincide with 5th order
fringe of 2nd source.
5. (d) Let P is a point infront of one slit at which intensity is to be
calculated. From figure,
Path difference = S
2
P – S
1
P
=  = 
= 
?x = 
Place difference,
?f = 
So, resultant intensity at the desired point ‘p’ is
I = I
0
cos
2
 = I
0
cos
2
6. (b) = 
and = .
As > , so shift will be along –ve y-axis.
7. (b) The width of the central maximum is given by
If  then ß decreases.
Also, intensity 
where                
I increases as d increases
The central maximum will become narrower and brighter.
8. (c) Here Angle of incidence, i = 57
tan 57° = 1.54
u
glass
 = tan i
It means, Here Brewster’s law is followed and the reflected ray is
completely polarised.
Now, when reflected ray is analysed through a polaroid then intensity of
light is given by malus law.
i.e. I = I
0
 cos
2
?
on rotating polaroid ‘?’ changes. Due to which intensity first decreases
and then increases.
9. (c) Here 
Q a
1
 = a
2
 = a
? 
Now,      ? 
.
10. (c) For path difference ?, phase difference= 2p rad.
For path difference , phase difference =  rad.
As K = 4I
0
  so intensity at given point where path difference is 
K' = 
= 
11. (c)
12. (c)
13. (b) For first minima at P
AP – BP = ?
AP – MP = 
So phase difference, f =  = p radian
14. (c) In Fraunhoffer diffraction, for minimum intensity, ?x 
For first minimum, m = 1
? ?x 
15. (a) As  and ,
? fringe width ß will decrease
16. (c) Distance  of nth maxima, 
As 
17. (a)
The intensity of light transmitted through third polaroid,
? intensity of light transmitted through the last polaroid
Page 5


1. (d) Optical path difference
?x = .
2. (b)
3. (a) For a circularly polarised light electric field remains constant with
time.
4. (b) Let nth fringe of 2500 Å coincide with (n – 2)th fringe of 3500Å.
?  3500 (n – 2) = 2500 × n
1000 n = 7000, n = 7
?  7th order fringe of 1st source will coincide with 5th order
fringe of 2nd source.
5. (d) Let P is a point infront of one slit at which intensity is to be
calculated. From figure,
Path difference = S
2
P – S
1
P
=  = 
= 
?x = 
Place difference,
?f = 
So, resultant intensity at the desired point ‘p’ is
I = I
0
cos
2
 = I
0
cos
2
6. (b) = 
and = .
As > , so shift will be along –ve y-axis.
7. (b) The width of the central maximum is given by
If  then ß decreases.
Also, intensity 
where                
I increases as d increases
The central maximum will become narrower and brighter.
8. (c) Here Angle of incidence, i = 57
tan 57° = 1.54
u
glass
 = tan i
It means, Here Brewster’s law is followed and the reflected ray is
completely polarised.
Now, when reflected ray is analysed through a polaroid then intensity of
light is given by malus law.
i.e. I = I
0
 cos
2
?
on rotating polaroid ‘?’ changes. Due to which intensity first decreases
and then increases.
9. (c) Here 
Q a
1
 = a
2
 = a
? 
Now,      ? 
.
10. (c) For path difference ?, phase difference= 2p rad.
For path difference , phase difference =  rad.
As K = 4I
0
  so intensity at given point where path difference is 
K' = 
= 
11. (c)
12. (c)
13. (b) For first minima at P
AP – BP = ?
AP – MP = 
So phase difference, f =  = p radian
14. (c) In Fraunhoffer diffraction, for minimum intensity, ?x 
For first minimum, m = 1
? ?x 
15. (a) As  and ,
? fringe width ß will decrease
16. (c) Distance  of nth maxima, 
As 
17. (a)
The intensity of light transmitted through third polaroid,
? intensity of light transmitted through the last polaroid
 
 = 
18. (a) Given geometrical spread = a
Diffraction spread 
The sum 
For b to be minimum 
b min = 
19. (d) Let any  its components are
with 
& 
there cos?
x
, cos?
y
 and cos?
z
 one called  direction cosines. 
Hence 
So, magnitude of 
and 
20. (a)
21. (500) Wavelength for which maximum obtained at the hole has the
maximum intensity on passing. So,