Problem no 1

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 1st
  minimum on the other side is 1.97º. Find the
  width of the slit.
• a sin ? m?
• 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 cma. calculate the angle of
diffraction of the second minimumb. find the
width of the slit

• a. sin?? d/D0.0162/2.16 7.5 x 10-3
• b. a sin ? m?
• On substituting
• a118 µ.m

3
Problem no 3

• A single slit is illuminated by light of
  wavelength are ?a and ?b so coherent that the
  first diffraction minimum of ?a component
  coincides with the second minimum of ?b
  component.
• A) what relationship exists between the two
  wavelengths
• B. Do any other minima in the two pattern
  coincide
• SOLUTION
• a sin? m?
• sin ? m?/a sin ?a1 sin ?b2
• 1?a/a 2?b/a ?a 2?b
• ma?a/a mb?b/a
• mb 2ma
• When ever mb is an integer ma is an even
  integer. i.e. All of the diffraction minima of ?a
  are overlapped by a minima of ?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? m?/a y/D
• y 2.02 mm

5
Problem no 5

• In a single slit diffraction pattern the distance
  between the 1st minimum on the right and the 1st
  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? m?/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 Newtons 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. rm2 mR? for Newton rings
• For m1, r12 ?R
• fm rm2 / m?
• For m1, f1 r12 / ? ?R/? R 1m
• B. Secondary focii put r22 and m2m-1 then we
  get
• R2R/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 m?/r m2
• For the central zone, m1, rm 1.15mm
• U500cm, ?5x10 -5 cm
• Hence v561.5 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? m?
• ? sin - ?/a (546x10 -9m)/2.58x10 -3 m
• 2.12x10 -4rad 1.21x10 -2 degree
• b) ?y/D (?m) ?/a
• a (?m) ?D/?y (5-1) (0.413) (546x10
  -9)/(0.35x10 -3)
• 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
• ? (m½)? where m1,2,3..
• I? Im sin (m½)?/ (m½)?2
• I? / Im 1/ (m½)?2
• 0.0450 for m1,
• 0.0162 for m2,
• 0.0083 for m3,
• 0.0050 for m4,
• 0.0033 for m5

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 ?D/d (480x10 -9) (52x10 -2)/(0.12x10-3)
  2.1mm
• Angular separation of the first minimum is
• sin? ?/a 0.0192
• Y D tan ? D sin? (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 (cos2ß) coincide
  with the 1st minimum of the diffraction factor
  (sin?/?)2.
• The sixth minimum of the interference factor
  occur when
• d sin? 11?/2 or ß 11?/2.
• The first minimum in the diffraction term occurs
  for dsin? ?
• Or ? ? and d/D 11/2 or d5.5

13
Problem no 11

• A. Design a double slit system in which the 4th
  fringe not counting the central maximum is
  missing.
• B. what other fringes if any are also missing?
• Soln
• A.
• d sin? 4? gives the location of the 4th
  interference maximum.
• a sin? ?, gives the location of the first
  diffraction minimum.
• If d 4a, there will be no 4th interference
  maximum.
• B.
• d sin?mi mmi? gives the location of the mth
  interference maxima.
• d sin?md mmd? gives the location of the m th
  diffraction minima
• D4a hence if m i 4md 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 observers eye to be
  4.6mm and the wavelength to be 542nm .
• Sol
• y/D 1.22?/a (here a4.6mm and y5.2mm)
• 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
• ?R sin -1(1.22?/a) 1.37 x 10-4 rad
• ?R y/D 1.42/D1.37x10 -4 rad.
• D1.04X 104

16
Diffraction grating problems

• A certain grating has 104 slits with a spacing
  d2100 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? m?
• sin? m (589 x 10 -9m)/(2100 x 10 -9m)
• For m 1, ?1 16.3
• For m 2, ?2 34.1
• For m 3, ?3 57.3
• For m 4, ?4 more than 90 degree hence 3.0
  order is the highest
• B) for m3, ?? ? / Nd cos? 5.2 x 10-5 rad or
  0.0030 degree

17
Grating contd

• A diffraction grating has 104 ruling uniformly
  spaced over 25 mm. It is illuminated normally
  using a sodium lamp containing two wavelengths
  589.0 and 589.59 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. ? sin-1 m?/d 13.6 degrees
• B. d? m??/ d cos ? 2.4 x 10-4 rads or 0.014
  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 104 rulings
  uniformly spaced over a width w 2.5 cm and is
  illuminated normally using sodium light
  containing two wavelengths 589 and 589.59nm.
• 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 ???
• still be resolved by this grating
• D. how many rulings can a grating have and just
  resolve the sodium doublet lines.
• soln
• A. ? sin-1 (m?/d) 16.4 degrees
• B. Dispersion D ??/?? m /(d cos? ) 5.0 x
  10-4 rad/nm
• ?? D x ?? 2.95 x
  10-4 rads or 0.0169 degrees
• C. Resolving power Nm 1.2 x 10 4
• 
  ?? ?/R 0.049 nm hence can resolve the D
  lines.
• D. R ?/?? 998. Hence no. of rulings needed
  is NR/m 998/1998 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
• ? (d sin?)/m 2367 nm for m1.
• On trying for m 4 5 we get in the visible range
  as 589nm and 469 nm and for m6 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
  ?1 tan -1 (50mm/300mm) 9.46 degree
• ? 1 (1 x 10 -3)sin 9.46 /350 470 nm
• The longest wavelength that can pass through an
  angle ?2 tan-1(60mm/300mm) 11.3 degree
• This corresponds to a wavelength
• ? 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 656.3 nm whose separation is
  0.180 nm. Find the minimum number of rulings
  needed in grating that can resolve these lines in
  the first order.
• SolnsN R/m ?/m?? 365

22
Grating contd

• A. How many rulings must a 4.15 cm wide
  diffraction grating have to resolve the
  wavelengths 415.496 nm and 415.487 nm in the
  second order.
• B. at what angle are the maxima found
• Soln
• NR/m ?/m?? 23100
• D w/N ..
• ? sin -1 m?/d 27.6 degrees