# Diffraction Ncert

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Get step by step NCERT solutions for Class 12 Physics Chapter 10 - Wave Optics. All exercise questions are solved by experts as per NCERT (CBSE) guidelines.

**Diffraction**

- Diffraction is the phenomenon by virtue of whichlight bendswhile passing through aslit or an opening.
- The extent of bending depends upon the diameter of the slit.
- Both Interference and Diffraction are closely related to each other.
- Young replaced the two slits by a single slit in his single slit experiment.Therefore this experiment is also referred as
__Young’s single slit experiment__. - When the light passed through one slit a different type of pattern was observed on the screen.
- The pattern which was observed had a central maximum band and which was very wide as compared to interference pattern.
- There were alternate dark and bright bands and their intensity was decreasing on both the sides.
- The central maxima,was very wide whereas corresponding secondary maxima and minima were reduced in the intensity.
- The change in the pattern was formed due to diffraction instead of interference.

__Conclusion:__-

- Reflection and refraction MCQ 1. A ray of light strikes the surface of mirror at an angle of 30 0 with the mirror.
- Diffraction data produced by Maurice Wilkins and Rosalind Franklin, proposed a very simple but famous Double Helix model for the structure of DNA. One of the hallmarks of their proposition was base pairing between the two strands of polynucleotide chains. However, this pr oposition was.

- Broad central bright band.
- Alternate dark and bright bands on either side.
- Intensity was decreasing on both sides.

__Diffraction Fringe Pattern__

- In diffraction there is an incoming wave which passes through a single slit and as a result diffraction pattern was obtained on the screen.

__Diffraction pattern Maxima__

- The incident wavefront is parallel to the plane of the slit. This shows they are in phase with each other.
- From the given figure:-
- Path difference =NP – LP
- =>NQ = a sinθ where a=width of the slit.
- The slit was divided into smaller parts M
_{1}, M_{2}, N and L, and when contribution from each part is added up. - At the central point θ =0. This implies path difference =0.
- This means all the parts of the slit will contribute completely. Therefore the intensity is the maximum.
**Central Maximum occurs at θ =0.**

** Problem:-**The light of wavelength 600nm is incident normally on a slit of width 3mm.Calcluate the linear width of central maximum on a screen kept 3m away from the slit?

__Answer:-__

Wavelength λ =600nm =600 x10^{-9}m.

Width of the slit a =3mm=3x10^{-3}m.

Distance of the screen D=3m.

Condition for minima:-a sinθ =nλ

=>a sin θ =λ

=>sin θ = (λ/a) = (600 x10^{-9})/ (3x10^{-3})

=>θ =(600 x10^{-9})/(3x10^{-3})

=>(x/D) = (600 x10^{-9})/ (3x10^{-3})

=> x= (600 x 10^{-9} x3)/ (3 x10^{-3})

x=600 x 10^{-6}m.

Therefore width =2x = 1200 x 10^{-6}m

Width = 1.2 mm

__Diffraction pattern Secondary Maxima__

Scientist Fresnel found that secondary maxima occurred when the value of θ is:

**θ =(n+(1/2)) (λ/a)**- For n=1 =>θ =(3λ)/(2a) = (1.5λ/a) (equation 1)
- Where = (3λ)/ (2a) it lies midway between 2 dark fringes.

- Suppose if the slit is divided into 3 parts ,
- Consider the first 2/3
^{rd }part of the slit , - Path difference = (2/3) a x θ
- =(2/3) a x(3λ/2a) using (equation 1)
- Path difference =λ.
- The λ is getting divide into 2 halves with path difference =λ/2 and λ/2.
- Each of λ/2 gets cancel with another λ/2.
- Contribution from (2/3
^{rd}) part gets cancel out with each other. - Therefore only (1/3
^{rd}) contributes to I(Intensity).- Intensity ≠ 0.
- Intensity is reduced.

- For n=2 => θ =(5 λ)/(a)
- Only (1/5
^{th}) part will contribute. - For n=3 =>θ =(7 λ)/(2a)
- Only (1/7
^{th}) part contributes.

- Consider the first 2/3

Slit divided into 3 parts

__Diffraction pattern for Minima__

- Condition to get minima on the diffraction pattern:
**θ = (n λ/a)**; n= (+-) 1, (+-2)…

- For n=1, =>θ = (λ/a)
- Suppose the slit is divided into small parts. For every M
_{1}in the portion LM there exists another M_{2}in MN. - From the figure it is clear that the contribution from all elements inLM (M
_{1}) will cancel out the contribution from all elements MN (M_{2}). - Therefore net intensity I=0 at θ = (λ/a).
- Minima occur at θ = (nλ/a). => aθ =nλ
- => a sinθ =nλ.
- Path difference = (nλ/a).
- Minima will occur at
**a sin θ =n λ.**Where a =width of the slit,λ = wavelength of the light used.

** Problem:- **A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtaininterference fringes in a Young’s double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the centralmaximum for wavelength 650 nm.

(b) What is the least distance from the central maximum where the bright fringes dueto both the wavelengths coincide?

__Answer:-__

Wavelength of the light beam, λ_{1} =650nm

Wavelength of another light beam,λ_{2} =520nm

Distance of the slits from the screen = D

Distance between the two slits = d

(a) Distance of the n^{th} bright fringe on the screen from the central maximum is given bythe relation,

x=nλ_{1}(D/d)

For third bright fringe, n=3

Therefore x=3x650(D/d) =1950(D/d) nm

(b) Let the n^{th} bright fringe due to wavelength and (n − 1)^{th} bright fringe due towavelength λ_{1}coincide on the screen. We can equate the conditions for bright fringesas:

nλ_{2}=(n-1)λ_{1}

520n=650n-650

650=130n

Therefore, n=5.

Hence, the least distance from the central maximum can be obtained by the relation:

Note: The value of d and D are not given in the question.

** Problem:- **In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screenplaced 1 m away. The wavelength of light used is 600 nm. What will be the angularwidthof the fringe if the entire experimental apparatus is immersed in water? Take refractiveindex of water to be 4/3.

__Answer:-__

Distance of the screen from the slits, D = 1 m

Wavelength of light used, λ_{1} =600nm

Angular width of the fringe in air θ_{1} =0.2^{o}

Angular width of the fringe in water = θ_{2}

Refractive index of water, μ= (4/3)

Refractive index is related to angular width as:

μ = (θ_{1}/θ_{2})

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θ_{2} = (3/4)θ_{1}

= (3/4)x0.2 =0.15

Therefore, the angular width of the fringe in water will reduce to 0.15°.

** Problem:- **A parallel beam of light of wavelength 500 nm falls on a narrow slitand the resulting diffraction pattern is observed on a screen 1 maway. It is observed that the first minimum is at a distance of 2.5mm from the centre of the screen. Find the width of the slit.

__Answer:-__

Wavelength of light beam, λ = 500 nm = 500 × 10^{−9} m

Distance of the screen from the slit, D = 1 m

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For first minima, n = 1

Distance between the slits = d

Distance of the first minimum from the centre of the screen can be obtained as:

x = 2.5 mm = 2.5 × 10^{−3} m

It is related to the order of minima as:

n λ =x(d/D)

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d= (n λD/x)

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d= (1x500x10^{-9}x1)/ (2.5 x10^{-3}) = 2x 10^{-4} m

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d=0.2mm

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Therefore, the width of the slits is 0.2 mm.