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Question : 25
Total: 26
(a) Using Huygens's construction of secondary wavelets explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a monochromatic beam of light is incident normally.
(b) Show that the angular width of the first diffraction fringe is half that of the central fringe.
(c) Explain why the maxima atθ = ( n +
)
n
OR
(a) A point object 'O n 1 R n 2
Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms ofn 1 , n 2 R
(b) When the image formed above acts as a virtual object for a concave spherical surface separating the mediumn 2 n 1 ( n 2 > n 1 )
(b) Show that the angular width of the first diffraction fringe is half that of the central fringe.
(c) Explain why the maxima at
OR
(a) A point object '
Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of
(b) When the image formed above acts as a virtual object for a concave spherical surface separating the medium
Solution:
(a) Explanation of diffraction pattern using Huygen's construction
(b) Showing the angular width of first diffraction fringe as half of the central fringe
(c) Explanation of decrease in intensity with increasingn
(a)
A plane wavefront from a monochromatic sourceS C C
The wavelets which meet at pointP P
Thus a diffraction pattern is generated.
(b)
Condition for first minimum on the screen
a s i n θ = λ
⇒ θ =
∴
= 2 θ =
Angular width of first diffraction fringe (From fig)
=
Hence angular width of central fringe is twice the angular width of first fringe.
(c) Maxima become weaker and weaker with increasingn n
OR
(a)
For small angles
∠ N O M = tan ∠ N O M =
∠ N C M = tan ∠ N C M =
∠ N I M = tan ∠ N I M =
In△ NOC ∠ i = ∠ N O M + ∠ N C M
∴ ∠ i =
+
Similarly,∠ r = ∠ N C M − ∠ N I M
=
−
Using Snell's Law
n 1 s i n i = n 2 s i n r
For small angles
n 1 i = n 2 r
Substituting fori r
+
=
Here,O M = − u , M I = + v , M C = + R
Substituting these, we get
⇒
−
=
Similarly relation for the surface ADC.
+
=
Refraction at the first surfaceA B C
+
=
Adding (i) and (ii) and takingB I 1 ≃ D I 1
+
= ( n 2 − n 1 ) (
+
)
Here,O B = − u
D I = + v
B C 1 = + R 1
⇒ D C 2 = − R 2
⇒
+
= − R 2
⇒ n 1 (
+
) = ( n 2 − n 1 ) (
+
)
⇒
−
) = (
− 1 ) (
−
)
(b) Showing the angular width of first diffraction fringe as half of the central fringe
(c) Explanation of decrease in intensity with increasing
(a)
A plane wavefront from a monochromatic source
The wavelets which meet at point
Thus a diffraction pattern is generated.
(b)
Condition for first minimum on the screen
Angular width of first diffraction fringe (From fig)
Hence angular width of central fringe is twice the angular width of first fringe.
(c) Maxima become weaker and weaker with increasing
OR
(a)
For small angles
In
Similarly,
Using Snell's Law
For small angles
Substituting for
Here,
Substituting these, we get
Similarly relation for the surface ADC.
Refraction at the first surface
Adding (i) and (ii) and taking
Here,
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