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Problem no 1  Light of wavelength 633 nm is incident on a narrow slit. The angle between the 1 st minimum on one side of the central maximum and the 1.

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Presentation on theme: "Problem no 1  Light of wavelength 633 nm is incident on a narrow slit. The angle between the 1 st minimum on one side of the central maximum and the 1."— Presentation transcript:

1 Problem no 1  Light of wavelength 633 nm is incident on a narrow slit. The angle between the 1 st minimum on one side of the central maximum and the 1 st minimum on the other side is 1.97º. Find the width of the slit.  a sin ө = mλ  a =633x10 -9 /sin(1.97/2)  36.8 micrometers

2 2. A monochromatic light of wavelength 441 nm falls on a narrow slit on a screen 2.16 away, the distance between the second and the central maximum is 1.62 cm a. calculate the angle of diffraction of the second minimum b. find the width of the slit  a. sin ө = ө =d/D=0.0162/2.16 =7.5 x 10-3  b. a sin ө = mλ  On substituting  a=118 µ.m

3 Problem no 3  A single slit is illuminated by light of wavelength are λ a and λ b so coherent that the first diffraction minimum of λ a component coincides with the second minimum of λ b component.  A) what relationship exists between the two wavelengths  B. Do any other minima in the two pattern coincide  SOLUTION:  a sin ө =mλ  sin ө =mλ/asin ө a1 = sin ө b2  1λ a /a = 2λ b /aλ a = 2λ b  m a λ a /a = m b λ b /a  m b =2m a  When ever m b is an integer m a is an even integer. i.e. All of the diffraction minima of λ a are overlapped by a minima of λ b

4 Problem no 4  A plane wave, with wavelength 593nm falls on a slit of width 420 µm. A thin converging lens having a focal length of 71.4 cm is placed behind the slit and focuses the light on a screen Find the distance on the screen from the center of the pattern to the second minimum  Solution:  sin ө = mλ/a = y/D  y = 2.02 mm

5 Problem no 5  In a single slit diffraction pattern the distance between the 1 st minimum on the right and the 1 st minimum on the left is 5.2 mm. The screen on which the pattern is displayed is 82.3 cm from the slit and the wavelength is 546 nm calculate the slit width. sin ө =mλ/a =y/D a = 173 micro meters

6 Zone plate problems  A zone plate is constructed in such a way that the radii of the circles which define the zones are the same as the radii of Newton’s rings formed between a plane surface and surface having radius o curvature of 2.0 m. a) Find the primary focal length of the zone plate and b) secondary focii  Soln;  A. r m 2 = mRλ for Newton rings  For m=1, r 1 2 = λR  f m = r m 2 / mλ  For m=1, f 1 =r 1 2 / λ = λR/λ = R =1m  B. Secondary focii; put r 2 2 and m=2m-1 then we get  R=2R/3= 0.66R m

7 Zone plate contd….  A point source of wavelength 5000A is placed 5.0 m away from the zone plate where central zone has the diameter 2.3 mm. Find the position of the primary image.  Soln:  1/f = i/u +i/v = mλ/r m 2  For the central zone, m=1, r m = 1.15mm  U=500cm, λ=5x10 -5 cm  Hence v=561.5 cm away from the zone plate

8 6. The distance between the first and the fifth minima of a single slit diffraction pattern is 0.350mm With the screen 41.3 cm away from the slit, using light of wavelength 546 nm A. calculate the diffraction angle of the first min B. find the width of the slit  Solution:  a) a sin ө =mλ ө =sin - λ/a =(546x10 -9 m)/2.58x10 -3 m =2.12x10 -4 rad =1.21x10 -2 degree  b) y/D = (m) λ/a  a = (m) λD/y = (5-1) (0.413) (546x )/(0.35x10 -3 ) =2.58 mm =2.58 mm

9 Problem no 7  If you double the width of a single slit, the intensity of the central maximum of the diffraction pattern increases by a factor of 4 times even through the energy passing through it only doubles. Explain qualitatively  Soln:  Doubling the width results in narrowing of the diffraction pattern As the width of the central maximum is effectively cut in half, then there is twice the energy in half the space, producing four times the intensity

10 Problem no 8  Calculate approximately the relative intensities of the maxima in the single slit,Fraunhofer Diffraction pattern  Soln:  The maxima lie app half way between the minima and are roughly given by   = (m=½) where m=1,2,3…..  I ө = I m {sin (m=½)/ (m=½)} 2  I ө / I m = {1/ (m=½)} 2  = for m=1,  = for m=2,  = for m=3,  = for m=4,  = for m=5

11 Problem no 9  In a double slit experiment the distance D of the screen from the slits is 52cm the wavelength is 480nm, the slit separation is 0.12mm and the slit width is 0.025mm  A.what is the spacing between adjacent fringes B.what is the distance from the cenetral maximum to the first minimum of the fringe envelope  Soln:  y = λD/d =(480x10 -9) ( 52x10 -2 )/(0.12x10 -3 ) =2.1mm  Angular separation of the first minimum is  sin ө =λ/a =  Y = D tan ө = D sin ө =(52x10 -2 )(0.0192) = 10mm  There are about 9 fringes in the central peak of of the diffraction envelope

12 Problem no 10  What requirements must be met for the central maximum of the envelope of the double slit interference pattern to contain exactly 11 fringes? How many fringes lie between the first and the second minima of the envelope?  Soln:  The required condition will be met if the 6 th min of the interference factor (cos 2 β) coincide with the 1 st minimum of the diffraction factor (sin/) 2. The sixth minimum of the interference factor occur when d sin ө = 11λ/2 or β = 11/2.  The first minimum in the diffraction term occurs for dsin ө = λ  Or  =  and d/D = 11/2 or d=5.5

13 Problem no 11  A. Design a double slit system in which the 4 th fringe not counting the central maximum is missing.  B. what other fringes if any are also missing?  Soln:  A.  d sin ө =4λ gives the location of the 4 th interference maximum.  a sin ө =λ, gives the location of the first diffraction minimum.  If d = 4a, there will be no 4 th interference maximum.  B.  d sin ө mi = m mi λ gives the location of the m th interference maxima.  d sin ө md = m md λ gives the location of the m th diffraction minima  D=4a hence if m i =4m d there will be a missing maxima

14 Problem no 12  The wall of large room is covered with acoustic tile in which small holes are drilled 5.2mm from the center to the center. How far can a person be from such a tile and still distinguish individual holes assuming ideal condition? Assume the diameter of the pupil of the observer’s eye to be 4.6mm and the wavelength to be 542nm.  Sol:  y/D = 1.22λ/a (here a=4.6mm and y=5.2mm)  D = 36.2m

15 Problem no 13  The two head lights of an approaching automobile are 1.42 m apart. At what  A) angular separation and  B) maximum distance will the eye resolve them?  Assume a pupil diameter of 5 mm and a wavelength of 562 nm. Also assume that the diffraction effects alone limit the resolution.  Solution:  A. least angular separation required for the resolution is  ө R = sin -1 (1.22λ/a) =1.37 x rad  ө R =y/D =1.42/D=1.37x10 -4 rad.  D=1.04X 10 4

16 Diffraction grating problems  A certain grating has 10 4 slits with a spacing d=2100 nm. It is illuminated with a light of wavelength 589 nm.  Find  A) The angular positions of all principal maxima observed and  B) the angular width of the largest order maximum.  Soln:  A. d sin ө = m λ  sin ө = m (589 x m)/(2100 x m)  For m = 1, ө 1 = 16.3  For m = 2, ө 2 = 34.1  For m = 3, ө 3 = 57.3  For m = 4, ө 4 = more than 90 degree hence 3.0 order is the highest  B) for m=3,  ө = λ / Nd cos ө = 5.2 x rad or degree

17 Grating contd……  A diffraction grating has 10 4 ruling uniformly spaced over 25 mm. It is illuminated normally using a sodium lamp containing two wavelengths and nm.  A. At what angle will the first order maximum occur for the first of these wavelengths?  B. what is the angular separation between the first order maxima for these lines. Will this alter in other orders.  A. ө = sin -1 m λ/d =13.6 degrees  B. d ө = m λ/ d cos ө =2.4 x rads or degrees.  As the spectral separation increases with the order no. this value increases with the order no.

18  A diffraction grating has 1.2 x 10 4 rulings uniformly spaced over a width w= 2.5 cm and is illuminated normally using sodium light containing two wavelengths 589 and nm.  A. at what angle does the first order maximum occur for the first of these wavelengths  B. what is the angular separation between these two lines in the first and the second orders  C. how close in wavelength can two lines be in the first order and ө λ  still be resolved by this grating  D. how many rulings can a grating have and just resolve the sodium doublet lines.  soln:  A.ө = sin -1 (m λ/d) = 16.4 degrees  B.Dispersion D =  ө/ λ = m /(d cos ө ) =5.0 x rad/nm  ө = D x λ =2.95 x rads or degrees  ө = D x λ =2.95 x rads or degrees  C.Resolving power = Nm = 1.2 x 10 4  λ =λ/R =0.049 nm hence can resolve the D lines.  D. R =λ/λ = 998. Hence no. of rulings needed is N=R/m =998/1=998 hence can easily resolve as it has 12 times no, of rulings in it.

19 Grating contd……  A grating has 200 ruling/mm and principal maximum is noted at 28 degrees. What are the possible wavelengths of the incident visible light  Soln:  λ = (d sin ө)/m = 2367 nm for m=1.  On trying for m =4 &5 we get in the visible range as 589nm and 469 nm and for m=6 and above it will be in the uv range.  

20 Grating contd……  For a grating the no. of rulings is 350/mm. A white light falling normally on it produces spectrum 30 cm from it. If a 10 mm square hole is cut in the screen with its inner edge 50mm from the central maximum and parallel to it, what range of wavelengths passes through the hole?  Soln:  Shortest wavelength passes through at an angle of ө 1 = tan -1 (50mm/300mm)= 9.46 degree  λ 1 ={ (1 x )sin 9.46} /350 = 470 nm  The longest wavelength that can pass through an angle ө 2 = tan - 1(60mm/300mm) = 11.3 degree  This corresponds to a wavelength  Λ 2 = {(1x10 -3 )sin11.3} / 350 = 560 nm 

21 Grating contd……  A source containing a mixture of hydrogen and deuterium atoms emit light containing two closely spaced red colors at nm whose separation is nm. Find the minimum number of rulings needed in grating that can resolve these lines in the first order.  Solns; N = R/m = λ/mλ =365

22 Grating contd……  A. How many rulings must a 4.15 cm wide diffraction grating have to resolve the wavelengths nm and nm in the second order.  B. at what angle are the maxima found  Soln:  N=R/m = λ/mλ =  D = w/N = …..  ө = sin -1 m λ/d = 27.6 degrees


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