Important Formulas: Wave Optics | Physics Class 12 - NEET PDF Download

 Page 1


WAVE OPTICS
Interference of waves of intensity ?
1
 and ?
2
 :
resultant intensity, ? = ?
1
 + ?
2
 + 
2 1
2 ? ? cos ( ? ?) where, ? ? = phase
difference.
For Constructive Interference : ?
max
 = 
? ?
2
2 1
? ? ?
For Destructive interference : ?
min
 = 
? ?
2
2 1
? ? ?
If sources are incoherent ? = ?
1
 + ?
2
 , at each point.
YDSE :
Path difference, ?p = S
2
P – S
1
P = d sin ?
if d < < D = 
D
dy
if y << D
for maxima,
?p = n ? ? y = n ? n = 0, ±1, ±2 .......
for minima
?p = ?p = 
?
?
?
?
?
?
?
?
?
?
?
?
?
3........ - 2, - -1, n
 2
) 1 n 2 (
.... 3......... 2, 1,  n 
2
) 1 n 2 (
?  y = 
?
?
?
?
?
?
?
?
?
?
?
?
?
3....... - 2, - -1, n
 2
) 1 n 2 (
.... 3......... 2, 1,  n 
2
) 1 n 2 (
where, fringe width ? = 
d
D ?
Here, ? = wavelength in medium.
Highest order maxima : n
max
 = ?
?
?
?
?
?
?
d
total number of maxima = 2n
max
 + 1
Highest order minima : n
max
 = ?
?
?
?
?
?
?
? 2
1 d
total number of minima = 2n
max
.
Page 2


WAVE OPTICS
Interference of waves of intensity ?
1
 and ?
2
 :
resultant intensity, ? = ?
1
 + ?
2
 + 
2 1
2 ? ? cos ( ? ?) where, ? ? = phase
difference.
For Constructive Interference : ?
max
 = 
? ?
2
2 1
? ? ?
For Destructive interference : ?
min
 = 
? ?
2
2 1
? ? ?
If sources are incoherent ? = ?
1
 + ?
2
 , at each point.
YDSE :
Path difference, ?p = S
2
P – S
1
P = d sin ?
if d < < D = 
D
dy
if y << D
for maxima,
?p = n ? ? y = n ? n = 0, ±1, ±2 .......
for minima
?p = ?p = 
?
?
?
?
?
?
?
?
?
?
?
?
?
3........ - 2, - -1, n
 2
) 1 n 2 (
.... 3......... 2, 1,  n 
2
) 1 n 2 (
?  y = 
?
?
?
?
?
?
?
?
?
?
?
?
?
3....... - 2, - -1, n
 2
) 1 n 2 (
.... 3......... 2, 1,  n 
2
) 1 n 2 (
where, fringe width ? = 
d
D ?
Here, ? = wavelength in medium.
Highest order maxima : n
max
 = ?
?
?
?
?
?
?
d
total number of maxima = 2n
max
 + 1
Highest order minima : n
max
 = ?
?
?
?
?
?
?
? 2
1 d
total number of minima = 2n
max
.
Intensity on screen : ? = ?
1
 + ?
2
 + 
2 1
2 ? ? cos ( ? ?) where, ? ? =
p
2
?
?
?
If ?
1
 = ?
2
,  ? = 4 ?
1
 cos
2
 ?
?
?
?
?
? ? ?
2
YDSE with two wavelengths ?
1
 & ?
2
 :
The nearest point to central maxima where the bright fringes coincide:
y = n
1
?
1
 = n
2
?
2
 = Lcm of ?
1
 and ?
2
The nearest point to central maxima where the two dark fringes
coincide,
y = (n
1
 – 
2
1
) ?
1
 = n
2
 – 
2
1
) ?
2
Optical path difference
?p
opt
 = ? ?p
? ? = 
?
? 2
 ?p = 
vacuum
2
?
?
  ?p
opt.
? = ( ? – 1) t. 
d
D
 = ( ? – 1)t 
?
B
.
YDSE WITH OBLIQUE INCIDENCE
In YDSE, ray is incident on the slit at an inclination of ?
0
 to
the axis of symmetry of the experimental set-up
     
?
1
P
1
P
2
B
0
O'
S
2
dsin ?
0
S
1
O
?
0
?
2
We obtain central maxima at a point where, ?p = 0.
or ?
2
 = ?
0
.
This corresponds to the point O’ in the diagram.
Hence we have path difference.
?p =
?
?
?
?
?
? ? ? ?
? ? ? ?
? ? ? ?
 O' below  points for ) sin d(sin
O' & O between points for ) sin (sin d
O above points for ) sin (sin d
0
0
0
... (8.1)
Page 3


WAVE OPTICS
Interference of waves of intensity ?
1
 and ?
2
 :
resultant intensity, ? = ?
1
 + ?
2
 + 
2 1
2 ? ? cos ( ? ?) where, ? ? = phase
difference.
For Constructive Interference : ?
max
 = 
? ?
2
2 1
? ? ?
For Destructive interference : ?
min
 = 
? ?
2
2 1
? ? ?
If sources are incoherent ? = ?
1
 + ?
2
 , at each point.
YDSE :
Path difference, ?p = S
2
P – S
1
P = d sin ?
if d < < D = 
D
dy
if y << D
for maxima,
?p = n ? ? y = n ? n = 0, ±1, ±2 .......
for minima
?p = ?p = 
?
?
?
?
?
?
?
?
?
?
?
?
?
3........ - 2, - -1, n
 2
) 1 n 2 (
.... 3......... 2, 1,  n 
2
) 1 n 2 (
?  y = 
?
?
?
?
?
?
?
?
?
?
?
?
?
3....... - 2, - -1, n
 2
) 1 n 2 (
.... 3......... 2, 1,  n 
2
) 1 n 2 (
where, fringe width ? = 
d
D ?
Here, ? = wavelength in medium.
Highest order maxima : n
max
 = ?
?
?
?
?
?
?
d
total number of maxima = 2n
max
 + 1
Highest order minima : n
max
 = ?
?
?
?
?
?
?
? 2
1 d
total number of minima = 2n
max
.
Intensity on screen : ? = ?
1
 + ?
2
 + 
2 1
2 ? ? cos ( ? ?) where, ? ? =
p
2
?
?
?
If ?
1
 = ?
2
,  ? = 4 ?
1
 cos
2
 ?
?
?
?
?
? ? ?
2
YDSE with two wavelengths ?
1
 & ?
2
 :
The nearest point to central maxima where the bright fringes coincide:
y = n
1
?
1
 = n
2
?
2
 = Lcm of ?
1
 and ?
2
The nearest point to central maxima where the two dark fringes
coincide,
y = (n
1
 – 
2
1
) ?
1
 = n
2
 – 
2
1
) ?
2
Optical path difference
?p
opt
 = ? ?p
? ? = 
?
? 2
 ?p = 
vacuum
2
?
?
  ?p
opt.
? = ( ? – 1) t. 
d
D
 = ( ? – 1)t 
?
B
.
YDSE WITH OBLIQUE INCIDENCE
In YDSE, ray is incident on the slit at an inclination of ?
0
 to
the axis of symmetry of the experimental set-up
     
?
1
P
1
P
2
B
0
O'
S
2
dsin ?
0
S
1
O
?
0
?
2
We obtain central maxima at a point where, ?p = 0.
or ?
2
 = ?
0
.
This corresponds to the point O’ in the diagram.
Hence we have path difference.
?p =
?
?
?
?
?
? ? ? ?
? ? ? ?
? ? ? ?
 O' below  points for ) sin d(sin
O' & O between points for ) sin (sin d
O above points for ) sin (sin d
0
0
0
... (8.1)
THIN-FILM INTERFERENCE
for interference in reflected light 2 ?d
= 
?
?
?
?
?
? ?
?
ce interferen ve constructi for )
2
1
n (
ce interferen e destructiv for n
for interference in transmitted light 2 ?d
= 
?
?
?
?
?
? ?
?
ce interferen e destructiv for )
2
1
n (
ce interferen ve constructi for n
Polarisation
? 
? ?  = tan   .(brewster's angle)
? ?  + ?
r
 = 90°(reflected and refracted rays are mutually
perpendicular.)
? Law of Malus.
I = I
0
 cos
2
I = KA
2
 cos
2
? Optical activity
? ?
C L
C
t
?
?
? ?
?
?
?  =  rotation in length L at concentration C.
Diffraction
? a sin ? = (2m + 1) /2  for maxima. where m = 1, 2, 3 ......
 ? sin ? =  
a
m ?
,  m = ? ?  1, ?  2, ?  3......... for minima.
? Linear width of central maxima =  
a
d 2 ?
? Angular width of central maxima = 
a
2 ?
Page 4


WAVE OPTICS
Interference of waves of intensity ?
1
 and ?
2
 :
resultant intensity, ? = ?
1
 + ?
2
 + 
2 1
2 ? ? cos ( ? ?) where, ? ? = phase
difference.
For Constructive Interference : ?
max
 = 
? ?
2
2 1
? ? ?
For Destructive interference : ?
min
 = 
? ?
2
2 1
? ? ?
If sources are incoherent ? = ?
1
 + ?
2
 , at each point.
YDSE :
Path difference, ?p = S
2
P – S
1
P = d sin ?
if d < < D = 
D
dy
if y << D
for maxima,
?p = n ? ? y = n ? n = 0, ±1, ±2 .......
for minima
?p = ?p = 
?
?
?
?
?
?
?
?
?
?
?
?
?
3........ - 2, - -1, n
 2
) 1 n 2 (
.... 3......... 2, 1,  n 
2
) 1 n 2 (
?  y = 
?
?
?
?
?
?
?
?
?
?
?
?
?
3....... - 2, - -1, n
 2
) 1 n 2 (
.... 3......... 2, 1,  n 
2
) 1 n 2 (
where, fringe width ? = 
d
D ?
Here, ? = wavelength in medium.
Highest order maxima : n
max
 = ?
?
?
?
?
?
?
d
total number of maxima = 2n
max
 + 1
Highest order minima : n
max
 = ?
?
?
?
?
?
?
? 2
1 d
total number of minima = 2n
max
.
Intensity on screen : ? = ?
1
 + ?
2
 + 
2 1
2 ? ? cos ( ? ?) where, ? ? =
p
2
?
?
?
If ?
1
 = ?
2
,  ? = 4 ?
1
 cos
2
 ?
?
?
?
?
? ? ?
2
YDSE with two wavelengths ?
1
 & ?
2
 :
The nearest point to central maxima where the bright fringes coincide:
y = n
1
?
1
 = n
2
?
2
 = Lcm of ?
1
 and ?
2
The nearest point to central maxima where the two dark fringes
coincide,
y = (n
1
 – 
2
1
) ?
1
 = n
2
 – 
2
1
) ?
2
Optical path difference
?p
opt
 = ? ?p
? ? = 
?
? 2
 ?p = 
vacuum
2
?
?
  ?p
opt.
? = ( ? – 1) t. 
d
D
 = ( ? – 1)t 
?
B
.
YDSE WITH OBLIQUE INCIDENCE
In YDSE, ray is incident on the slit at an inclination of ?
0
 to
the axis of symmetry of the experimental set-up
     
?
1
P
1
P
2
B
0
O'
S
2
dsin ?
0
S
1
O
?
0
?
2
We obtain central maxima at a point where, ?p = 0.
or ?
2
 = ?
0
.
This corresponds to the point O’ in the diagram.
Hence we have path difference.
?p =
?
?
?
?
?
? ? ? ?
? ? ? ?
? ? ? ?
 O' below  points for ) sin d(sin
O' & O between points for ) sin (sin d
O above points for ) sin (sin d
0
0
0
... (8.1)
THIN-FILM INTERFERENCE
for interference in reflected light 2 ?d
= 
?
?
?
?
?
? ?
?
ce interferen ve constructi for )
2
1
n (
ce interferen e destructiv for n
for interference in transmitted light 2 ?d
= 
?
?
?
?
?
? ?
?
ce interferen e destructiv for )
2
1
n (
ce interferen ve constructi for n
Polarisation
? 
? ?  = tan   .(brewster's angle)
? ?  + ?
r
 = 90°(reflected and refracted rays are mutually
perpendicular.)
? Law of Malus.
I = I
0
 cos
2
I = KA
2
 cos
2
? Optical activity
? ?
C L
C
t
?
?
? ?
?
?
?  =  rotation in length L at concentration C.
Diffraction
? a sin ? = (2m + 1) /2  for maxima. where m = 1, 2, 3 ......
 ? sin ? =  
a
m ?
,  m = ? ?  1, ?  2, ?  3......... for minima.
? Linear width of central maxima =  
a
d 2 ?
? Angular width of central maxima = 
a
2 ?
?
2
0
2 /
2 / sin
?
?
?
?
?
?
?
?
? ? ?
  where  ?  =  
?
? ? sin a
? Resolving power .
R =  
? ?
?
?
? ?
?
1 2
–
where ,  
2
2 1
? ? ?
? ? , ? ?   =  ?
2
 -  ?
1