1.

Transverse wave nature

2. diffraction

3. photoelectric effect

4. They produce wave of same wavelength having a constant initial phase difference

5. conservation of energy

6. (2n-1)π

7. n λ
𝜆
8. (2n-1)
2

9. photoelectric effect
𝐷
10. β =𝑑 𝜆

11. Energy is neither created nor destroyed

12. must be monochromatic

13. Bright

14. their phase difference remains constant

15.Decreaes

16. 2π
𝑛𝜆𝐷
17.
𝑑

18. Bright

19.

H int,
d  2 x    1 
D
and Use  =
d
20.

f thin plate is introduced

D
x=    1 t
d
D
 
d
 D

 d

 x=    1 t


21. 10-4mm

22. plane
23. Area of Half Zone = πa 𝜆
𝑟𝑛2
Focal length =
𝑛𝜆

24. For minimum

dsin ϴ = n 𝜆

25. Hint dsin ϴ = n 𝜆

1
26. (a+b) = 𝑁, For Maximum (a+b) sin ϴ = n 𝜆

27. For missing order

(a+b) sin ϴ = n 𝜆

If 𝜆 𝑎𝑛𝑑 ϴ are so adjusted such that

asin ϴ = p 𝜆

On Dividing both

ab n

a p
 b  d
n  p 1    p 1  
 a  
if d=
 n=2p
1.22𝜆
28. Resolving power dϴ = 𝑑
𝐿
29. -𝑐
𝐷
30. β =𝑑 𝜆
𝐵
µ = A + 𝜆2

if VIBGYOR

Red has large wavelength

µb> µr
1
ß⋉µ

Difraction become narrow and crowded together

31.

x
ϴ

D
 x
ϴ= 𝐷

dsinϴ = 𝜆
𝜆
ϴ=
𝑑
𝐷
β =𝑑 𝜆 i.e
𝐷 β
=
𝑑 𝜆

For Central Maximum

2 ϴ = 0.36

32. For minimum,

dsinϴ = n𝜆

33. π
𝐷
34. Hint β =𝑑 𝜆, 𝐼𝑛𝑐𝑟𝑒𝑎𝑒𝑠

35. A cheerleader yells through a megaphone

36. Refraction

37. the phase angle between the extreme rays

𝐷
38. β =𝑑 𝜆 𝑖. 𝑒 𝑖𝑓 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔ℎ𝑡 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑒𝑛 𝑠𝑙𝑖𝑡 𝑤𝑖𝑑𝑡ℎ

39. The width of the pattern on the screen at first decreases but then increases

40. π

41. 1.1 rad

42. dsinϴ = 𝑛𝜆
3𝜆
Slightly less than 2𝑎 ,

43. Some Huygens wavelets sum to zero at the secondary maximum but not at the central maximum
𝐷
44. β =𝑎 𝜆
1.22𝜆 𝑥
45. dϴ = 𝑑
, dϴ = 𝐷

1.22𝜆 𝑥
∴ 𝑑
=𝐷
1.22𝜆
46. dϴ =
𝑑

1 𝑑
Resolving power = dϴ = 1.22𝜆

47. Becomes narrower

1.22𝜆
48. dϴ = 𝑑
 dϴ ⋉ 𝜆
1
dϴ ⋉ d

∴d&𝜆
49. (a+b)sin ϴ = 2 𝜆
1
50. (a+b) = 10000

And then use,

(a+b)sin ϴ = 𝜆
1
51. ß⋉ µ

52. Here,

(a + b ) sin ϴ = 2 𝜆

𝑤
And use (a + b) =
𝑁

( 𝑎+𝑏 ) 𝑛
53. 𝑎
= 𝑝

If a = b

n = 4p

( 𝑎+𝑏 ) 𝑛
54. Hint . =
𝑎 𝑝

55. Half angular width of principal maxima

It is defined as the angular width of principal maxima at

which intensity of light falls to onehalf of peak value

 1.4 
Half angular width  =2sin 1  
 a 
56. Dispersion of
𝜆
Diffraction Grating 𝑑𝜆= nN
1
Resolving Power = , dϴ is limit of resolution
𝑑𝛳
𝑑𝛳 𝑛
Dispersive power, 𝑑𝜆 = 𝑎+𝑏𝑐𝑜𝑠𝛳

Separation of lines of same order

Resolving power independent of Grating Element

Note
 A Grating with High dispersive power then
another doesnot necessirily have high
resolving power
-with increase in no. of lines per cm,
Dispersive power increase
 with increase in total no. of lines in
Effective part of grating, Resolving
power increase
2.54
57. (a+b) = 18000 per cm

58. Hint,

Area of zone = πa𝜆

𝜋 𝑟2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎𝑝𝑒𝑟𝑡𝑢𝑟𝑒
n = πb𝜆 = Area of half period zone

𝑑2
f = 𝑛𝜆

OR In Summary,

rn 2
f=
n
r2
n
b
r  fn
2

59. Hint ,

Area of n th  na
 r2 r2
Circular aperture n= 
 b b
Focal length

60. Phase

61. to create path length difference

𝐷
62. β =𝑑 𝜆 , ℎ𝑒𝑟𝑒 𝑑 = 2𝑑

63. Bright and Dark fringes alter

𝜆
64. 4

65. Hint Use

𝜆
2 µtcosr = (2n+1) 2

As i=o. r=0 Then cosr=1

As per question,

𝑛1 𝜆1 = 𝑛2 𝜆2 , 𝑛2 = 𝑛1 + 1
𝜆
66. Hint use, ß =
2µ𝛳

ß1 4
ß2
= 3

And Then use , ß1 − ß2

67.

t1 t2
ϴ
x1

x2
𝜆 𝜆
ß= 2µ𝛳
, t = 2µ

Diff of film thickness between two points

𝑡1 𝑡1
X2 = X1 =
𝛳 𝛳

𝑡2 −𝑡1 𝑥2 −𝑥1 𝜆
X2-x1 = 𝛳
, Here Fringe width ß = 𝑚
= 2µ𝛳

68. Hint Imax = (a1+a2)2

. Imin = (a1-a2)2

69. Let amplitude proportional to width

𝐴1 4
𝐴2
=9

4
(A1`+A2 ) = {𝐴2 9 + 𝐴2 }
13 5
= 𝐴
9 2
and (A1`- A2 ) = 𝐴
9 2

Finally,
𝐼𝑚𝑎𝑥 (A1`+ A2 ) 2
𝐼𝑚𝑖𝑛
= { (A1`− A2 ) }

70. VIBGOYR

𝜆𝑟> 𝜆𝑏
𝜆𝑦> 𝜆𝑏

µ𝑏> µ𝑦
1
ß⋉
µ
ßy> ßb

Remains unchanged
𝜆𝐷
71. Use , ß = and subtract ß2- ß1
𝑑

72. Hint use d = √𝐶1 𝐶2

73. Equidistance alternate dark and bright straight line fringes

74. Dark

75. Newton’s ring,

𝑟2
2t = 𝑅

r = √𝑛𝜆𝑅,

4 𝑅𝑛𝜆
dn2 = µ

4 𝑅𝑛𝜆
76. dn2 = µ

77.

12 R
rnth dark ring=
1  2

78. If m1 & m2 are perpendicular , Circular fringes are produced.

79.


n l
2
2l

n
For Refractive index
2    1 t  n
for l=n

2    1 t  l

D
and x=    1 t
d
80. Llyod’s Mirror
𝜆𝐷
ß= 𝑑

1.22𝜆
81 dϴ = 𝑑
 1
Resolving Power = 𝑑𝛳

82 microscope
1.22 𝜆
X = 2 sin 𝑖 , if 𝜆 𝑖𝑠 𝑖𝑛 𝑙𝑖𝑞𝑢𝑖𝑑 𝑎𝑛𝑑 𝜆0 𝑖𝑛 𝑣𝑎𝑐𝑐𝑢𝑚
𝜆0
𝜆= µ

𝜆
83. 𝑑𝜆 = 𝑛𝑁
1.22𝜆
84. Hint dϴ =
𝑑

𝑥
dϴ=𝑑

85. √𝜆𝐷

86. 𝜆1 = 5896 𝐴

𝜆2 = 5890 𝐴

Mean wavelength = 5893

0.5
(a+b) = 2500

If ϴ is mean angle of diffraction for sodium line in first order, then

𝜆
sinϴ= , cosϴ = √1 − 𝑠𝑖𝑛2 𝛳
𝑎+𝑏

𝑛𝑑𝜆
Angular separation dϴ = (𝑎+𝑏)𝑐𝑜𝑠𝛳

Two lines will be seen distinctly if they are resolved by grating,

𝜆
∴ Reqd resolving power = 𝑑𝜆

Actual Resolving power = nN

Finally (a+b)sinϴ = n𝜆

cos𝛳𝑑𝛳(𝑎 + 𝑏) = 𝑛 𝑑𝜆
𝑛𝑑𝜆
dϴ = (𝑎+𝑏)𝑐𝑜𝑠𝛳

87.

 2N
d
88. mean wavelength = 400.05
 1
N=
d 2
D𝜆 = 0.1
𝜆
. 𝑑𝜆
=nN

𝜆
89. Hint . 𝑑𝜆
=nN
𝑤𝑖𝑑𝑡ℎ
90 Hint a+b = 𝑛𝑜

𝜆
. 𝑑𝜆
=nN
𝜆𝐷
91. ß = 𝑑

𝜆𝐷
92. ß =
𝑑

𝑑ß
93. D =
𝜆

𝜆𝐷
94. Hint ß = 𝑑
𝑎𝑛𝑑 𝑢𝑠𝑒 ß2 − ß1

95. the crest from one wave overlaps with the trough from the other

96. dsin ϴ = n 𝜆

97. behaves like wave

98. 4 𝜆

99. Hint, use dsin ϴ1 = n 𝜆 and dsin ϴ2 = n 𝜆 Divide both

100. For 4th Order,

3 down , 3 up + 1 maximum

3+3+1=7

101. Interference

102. Phase shift of light when reflected

103.

104.

105.

106. For transmitted, Constructive

2 µt = n 𝜆

108. For Relection

𝜆
2 µt = (2𝑛 + 1)
2

109. 0

µ2>µ1 & µ3>µ2

π + π = 2π

110. Single Slit

𝑥 𝑦
111. sinϴ = 𝑑, tanϴ = 𝐷
𝑦
For Small angle sinϴ ≈ tanϴ =
𝐷

ß
Here, ϴ = 𝐷
𝜋 𝐷 1
2.2 × 180 = 𝜆 𝑑 𝐷

112. Follow 111.

𝜆
113. sinϴ = 𝑑

114. Hint, (a +b)sin ϴ = n𝜆

115. , (a +b)sin ϴ = n𝜆

116. , (a +b)sin ϴ = n𝜆

117. , (a +b)sin ϴ = n𝜆

118. , (a +b)sin ϴ = n𝜆
𝜆 1
119. (a+b) = 𝑠𝑖𝑛ϴ and use N = (𝑎+𝑏)

220. Use , (a +b)sin ϴ = n𝜆

121. ϴ1 = 24.30; ϴ2=55.38

122. ϴ1 = 10.66; ϴ2=16.11

123. Neither screen nor source at infinite distance from slit

124. For Diffraction: Narrow slit is important

125. interference minima are perfectly dark but that of diffraction may not be Dark
𝜆𝐷
126. ß = 𝑑

𝜆𝐷
127. Hint, ß = 𝑑

128. Ram First I visit u at Xavier Graduation Ceremony

Radio>micro>Infra>Visible>Ultra>X-ray>y-ray>Cosmic
𝜆𝐷
ß= µ𝑑

1
i.e µ ∝
𝜆

129. Wave nature

𝜆𝐷
130. ß = 𝑑
, 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠

131. Greater for narrow slit

132. Small as diameter of aperture is large

𝜆
133. sinϴ = 𝑑

134. Becomes narrower

135. Diffraction bands become narrower and crowded together

136. Should be same order of wavelength

137. Red
𝜆
138. = 𝑛𝑁
𝑑𝜆

139. (a+b) ∝ 𝑛

140. Dispersive Power

𝑑𝑦 𝑛 𝑡𝑎𝑛ϴ
𝑑𝑥
= (𝑎+𝑏)𝑐𝑜𝑠ϴ
= 𝜆
, not directly proportional to wavelength

𝜆
141. sinϴ = 𝑑

142. wider

143. Plane

144. Interference

145. Biprism ( Narrow Slit )

146. rn = √𝑛𝜆𝑅, dn ∝ √𝑛
147. Transmitted
148. Bright and dark fringes will alternate
149. Refraction
𝜆𝐷
150. ß = 𝑑
2𝜋
151.
𝜆
152. Vertically Downwards slightly
Point to note: (Newton Rings)
If lens slightly move upwards
Center becomes alternatively dark & bright with larger & larger
radius of central ring.
153. Wavelength and const phase diff
154. 𝜋
155. Both
 𝜆𝐷
156. use ß = 𝑑
𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 ß5 − ß3
157. n𝜆
𝜆𝐷
158. ß = 𝑑µ
159. should be of orderof wavelength of light used.
𝜆𝐷
160 ß = 𝑑
161. Let amplitude proportional to width

𝐴1 1
𝐴2
=9
𝐼𝑚𝑎𝑥 10𝐴1 2
=( )
𝐼𝑚𝑖𝑛 8𝐴1
𝐼 √𝐼1 +√𝐼2
162. 𝐼𝑚𝑎𝑥 = { }
𝑚𝑖𝑛 √𝐼1 −√𝐼2

163. 10,000A
164. Location of observer
165. Redistributed
166. 1 µm
167. Here
I max   a11  a2 
2

I min   a1  a2 
2

let a1  a2
Then
I max  4a12
But
I max a 2
Then I max  4 I 0

Imax− Imin
168. Fringe Contrast = Imax+ Imin

2𝑎 𝑎
Visiblity = FC = 𝑎2 +1 𝑎22
1 2
1.22𝜆
169. dϴ = 𝑑

𝑥
dϴ = 𝐷
1
170. Resolving power of microscope = 𝑑ϴ
1.22𝜆
x = 2µ𝑠𝑖𝑛𝑖, l
limited by diameter of objective lens

171. Divergence
172. 3m
173. Rayleigh “Two illuminated objects lying close to each other are said to
resolved if centrl maximum diffraction pattern of one falls on first 2 minimum of other.”
174. Diameter of objective lens
175. Diameter of objective lens
176. Resolving power of prism
 𝜆 𝑑µ
𝑑𝜆
= 𝑑𝜆
Directly proportional to rate of change of RI with wavelength

1 𝑑
177. Resolving power = = , 𝜆𝑏 < 𝜆𝑟 , Increases
𝑑ϴ 1.22 𝜆

𝑑
178. Resolving power = ,
1.22 𝜆
179. Diffraction of light
1
180. Resolving power ∝
𝜆1
𝑑
181. Resolving power = 1.22 𝜆
, , To increase RP
Both focal length and aperture of objective has to be increased.
𝑑
182. RP = Resolving power = 1.22 𝜆,
1.22𝜆
183. dϴ = 𝑑
1 𝑑
184. Resolving limit = = ,
𝑑ϴ 1.22 𝜆
185. Directly on objective lens
186.
Here,
Here
10  60 min
 1 
1 min=  
 60 
 1 
1sec =  
 3600 
Here,
  1 
   radian
 180 60 
3
d 
x
3
x  10 Km
d

187. High Resolving Power

Tips

There are 60 minutes per degree & 60 second per

Minute & 180 degree in pi radia

7 3
For eg: 70 d 7m 30s = {70 + 60 + 3600}
 Note :
 Two Telescope with objective of same
Diameter will have diff magnifying
power
 In diffraction grating if,
 >a+b, not possible
. all  a  b  sin   
 If plane glass is used instead of
glass in  NRS  , fringes will Disappear
 Areial Speed=  d f 
2

Power
Intensiry 
Area