Diffraction Grating Formula - General Physics - Solved Exam, Exams of Physics

This is the Solved Exam of General Physics which includes Moment of Force, Set of Co-Planar Forces, Maximum Angular Velocity, Hooke’s Law, Frequency of Vibration, Equilibrium Position etc. Key important points are: Diffraction Grating Formula, Diffraction and Dispersion, Interference Path Difference, Wavelength of Green Light, Maximum Number of Images, Central Image, Ray of Light, Lower Refractive Index

Typology: Exams

2012/2013

Uploaded on 02/19/2013

pankita
pankita 🇮🇳

4.5

(15)

82 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
1
2009 Question 7.
When light shines on a compact disc it acts as a diffraction grating causing diffraction and dispersion of
the light.
(i) Explain diffraction
Diffraction is the spreading out of a wave when it passes through a gap or passes by
an obstacle.
(ii) Explain dispersion.
Dispersion is the splitting up of white light into its constituent colours.
(iii)Derive the diffraction grating formula.
For constructive interference path difference = nλ, where n is an integer
From diagram we can see that path difference = d sin θ
nλ = d sin θ
(iv) An interference pattern is formed on a screen when green light from a laser passes normally through a
diffraction grating. The grating has 80 lines per mm and the distance from the grating to the screen is 90
cm. The distance between the third order images is 23.8 cm.
Calculate the wavelength of the green light.
d = 1/80000 = 1.25 × 10-5 m
θ = tan-1 (0.238/0.90)
n = 3
nλ = d sin θ λ = d sin θ/n λ = 551 (± 5) × 10-9 m.
(v) Calculate the maximum number of images that are formed on the screen.
For maximum number θ = 900 sin θ = 1
nλ = d sin θ nλ = d  n= d/λ
n = 22.7 so the greatest whole number of images is 22.
But this is on one side only.
In total there will be 22 on either side, plus one in the middle, so total = 45
(vi) The laser is replaced with a source of white light and a series of spectra are formed on the screen.
Explain how the diffraction grating produces a spectrum.
Different colours have different wavelengths so constructive interference occurs at different positions for
each separate wavelength.
(vii) Explain why a spectrum is not formed at the central (zero order) image.
At central image θ = 0 so constructive interference occurs for all separate wavelengths at the same point
so no separation of colours.
pf2

Partial preview of the text

Download Diffraction Grating Formula - General Physics - Solved Exam and more Exams Physics in PDF only on Docsity!

1

2009 Question 7. When light shines on a compact disc it acts as a diffraction grating causing diffraction and dispersion of the light. (i) Explain diffraction Diffraction is the spreading out of a wave when it passes through a gap or passes by an obstacle. (ii) Explain dispersion****. Dispersion is the splitting up of white light into its constituent colours. (iii) Derive the diffraction grating formula. For constructive interference path difference = nλ, where n is an integer From diagram we can see that path difference = d sin θ nλ = d sin θ (iv) An interference pattern is formed on a screen when green light from a laser passes normally through a diffraction grating. The grating has 80 lines per mm and the distance from the grating to the screen is 90 cm. The distance between the third order images is 23.8 cm. Calculate the wavelength of the green light. d = 1/80000 = 1.25 × 10-5^ m θ = tan-1^ (0.238/0.90) n = 3 nλ = d sin θ  λ = d sin θ/n  λ = 551 (± 5) × 10-9^ m. (v) Calculate the maximum number of images that are formed on the screen. For maximum number θ = 90^0  sin θ = 1 nλ = d sin θ  nλ = d  n= d/λ n = 22.7 so the greatest whole number of images is 22. But this is on one side only. In total there will be 22 on either side, plus one in the middle, so total = 45 (vi) The laser is replaced with a source of white light and a series of spectra are formed on the screen. Explain how the diffraction grating produces a spectrum. Different colours have different wavelengths so constructive interference occurs at different positions for each separate wavelength. (vii) Explain why a spectrum is not formed at the central (zero order) image. At central image θ = 0 so constructive interference occurs for all separate wavelengths at the same point so no separation of colours.

2

2009 Question 12 (c) Information is transmitted over long distances using optical fibres in which a ray of light is guided along a fibre. Each fibre consists of a core of high quality glass with a refractive index of 1.55 and is coated with glass of a lower refractive index. (i) Explain, with the aid of a labelled diagram, how a ray of light is guided along a fibre.

  1. An optical fibre consists of a glass pipe coated with a second material of lower refractive index.
  2. Light enters one end of the fibre and strikes the boundary between the two materials at an angle greater than the critical angle, resulting in total internal reflection at the interface.
  3. This reflected light now strikes the interface on the opposite wall and gets totally reflected again.
  4. This process continues all along the glass pipe until the light emerges at the far end. (ii) Why is each fibre coated with glass of lower refractive index? Because total internal reflection can only occur for rays travelling from a denser to a rarer medium. (iii) What is the speed of the light as it passes through the fibre? n = cair/cglass  cglass = 3.0 × 10^8 /1. cglass = 1.94 × 10^8 m s- (iv) Light passing through optical fibres must travel through an enormous length of glass. Impurities in the glass reduce the power transmitted by half every 2 km. The initial power being transmitted by the light is 10 W. What is the power being transmitted by the light after it has travelled 8 km through the fibre? After 2 km power has dropped to 5 W; after 4 km power has dropped to 2.5 W; after 6 km power has dropped to 1.25 W; after 8 km power has dropped to 0.625 W.