Problem no 1Problem no 1

Light of wavelength 633 nm is Light of wavelength 633 nm is incident on a narrow slit . The angle incident on a narrow slit . The angle between the 1 between the 1 stst minimum on one side minimum on one side of the central maximum and the 1of the central maximum and the 1stst minimum on the other side is 1.97minimum on the other side is 1.97º. º. Find the width of the slit.Find the width of the slit.

a sin a sin өө = m = mλλ a =633x10a =633x10-9-9/sin(1.97/2)/sin(1.97/2) 36.8 micrometers36.8 micrometers

2.2. A monochromatic light of wavelength 441 nm falls A monochromatic light of wavelength 441 nm falls on a narrow slit on a screen 2.16 away, the distance on a narrow slit on a screen 2.16 away, the distance between the second and the central maximum is 1.62 between the second and the central maximum is 1.62 cmcma. calculate the angle of diffraction of the second a. calculate the angle of diffraction of the second minimumminimumb. find the width of the slit b. find the width of the slit

a. sina. sinөө==өө =d/D=0.0162/2.16 =7.5 x 10-3 =d/D=0.0162/2.16 =7.5 x 10-3 b. a sin b. a sin өө = m = mλλ On substitutingOn substituting a=118 µ.m a=118 µ.m

Problem no 3Problem no 3 A single slit is illuminated by light of wavelength are A single slit is illuminated by light of wavelength are λλaa and and

λλb b so coherent that the first diffraction minimum of so coherent that the first diffraction minimum of λλaa component coincides with the second minimum of component coincides with the second minimum of λλb b component. component.

A) what relationship exists between the two wavelengths A) what relationship exists between the two wavelengths B. Do any other minima in the two pattern coincideB. Do any other minima in the two pattern coincide

SOLUTION:SOLUTION: a sina sinөө =m =mλλ sin sin өө =m =mλλ/a/a sin sin өөa1a1 = sin = sin өөb2b2 11λλaa/a = 2/a = 2λλbb/a/a λλaa = 2 = 2λλbb mmaaλλaa/a = m/a = mbbλλbb/a/a mmb b =2m=2maa When ever mWhen ever mbb is an integer m is an integer maa is an even integer. i.e. All of the is an even integer. i.e. All of the

diffraction minima of diffraction minima of λλa a are overlapped by a minima of are overlapped by a minima of λλbb

Problem no 4Problem no 4 A plane wave, with wavelength A plane wave, with wavelength

593nm falls on a slit of width 420 593nm falls on a slit of width 420 µµm. m. A thin converging lens having a focal A thin converging lens having a focal length of 71.4 cm is placed behind length of 71.4 cm is placed behind the slit and focuses the light on a the slit and focuses the light on a screen Find the distance on the screen Find the distance on the screen from the center of the pattern screen from the center of the pattern to the second minimum to the second minimum

Solution:Solution: sinsinөө = m = mλλ/a = y/D/a = y/D y = 2.02 mmy = 2.02 mm

Problem no 5Problem no 5

In a single slit diffraction pattern the In a single slit diffraction pattern the distance between the 1distance between the 1st st minimum on minimum on the right and the 1the right and the 1stst minimum on the minimum on the left is 5.2 mm. The screen on which left is 5.2 mm. The screen on which the pattern is displayed is 82.3 cm the pattern is displayed is 82.3 cm from the slit and the wavelength is from the slit and the wavelength is 546 nm calculate the slit width.546 nm calculate the slit width.sinsinөө =m =mλλ/a =y/D/a =y/D

a = 173 micro metersa = 173 micro meters

Zone plate problemsZone plate problems A zone plate is constructed in such a way that the A zone plate is constructed in such a way that the

radii of the circles which define the zones are the radii of the circles which define the zones are the same as the radii of Newton’s rings formed between same as the radii of Newton’s rings formed between a plane surface and surface having radius o a plane surface and surface having radius o curvature of 2.0 m. a) Find the primary focal length curvature of 2.0 m. a) Find the primary focal length of the zone plate and b) secondary fociiof the zone plate and b) secondary focii

Soln;Soln; A. rA. rmm

22 = mR = mRλ λ for Newton ringsfor Newton rings For m=1, rFor m=1, r11

2 = 2 = λλRR ffmm = r = rmm

2 2 / m/ mλλ For m=1, fFor m=1, f11 =r =r11

2 2 / / λλ = = λλR/R/λλ = R =1m = R =1m B. Secondary focii; put rB. Secondary focii; put r22

22 and m=2m-1 then we get and m=2m-1 then we get R=2R/3= 0.66R mR=2R/3= 0.66R m

Zone plate contd….Zone plate contd…. A point source of wavelength 5000A A point source of wavelength 5000A

is placed 5.0 m away from the zone is placed 5.0 m away from the zone plate where central zone has the plate where central zone has the diameter 2.3 mm. Find the position of diameter 2.3 mm. Find the position of the primary image.the primary image.

Soln:Soln: 1/f = i/u +i/v = m1/f = i/u +i/v = mλλ/r /r mm

22 For the central zone, m=1, rFor the central zone, m=1, rmm = 1.15mm = 1.15mm U=500cm, U=500cm, λλ=5x10 =5x10 -5-5 cm cm Hence v=561.5 cm away from the zone plate Hence v=561.5 cm away from the zone plate

6. The distance between the first and the fifth minima 6. The distance between the first and the fifth minima of a single slit diffraction pattern is 0.350mm With the of a single slit diffraction pattern is 0.350mm With the screen 41.3 cm away from the slit, using light of screen 41.3 cm away from the slit, using light of wavelength 546 nm wavelength 546 nm A. calculate the diffraction angle of the first min A. calculate the diffraction angle of the first min B. find the width of the slit B. find the width of the slit

Solution:Solution: a) a a) a sinsinөө =m =mλλ

өө =sin =sin - - λλ/a =(546x10 /a =(546x10 -9-9m)/2.58x10 m)/2.58x10 -3 -3 mm

=2.12x10 =2.12x10 -4-4rad =1.21x10 rad =1.21x10 -2-2 degree degree b) b) y/D = (y/D = (m) m) λλ/a/a a = a = ((m) m) λλD/D/y =y =(5-1) (0.413) (546x10(5-1) (0.413) (546x10 -9 -9)/(0.35x10 )/(0.35x10 --

33)) =2.58 mm=2.58 mm

Problem no 7Problem no 7 If you double the width of a single If you double the width of a single

slit, the intensity of the central slit, the intensity of the central maximum of the diffraction pattern maximum of the diffraction pattern increases by a factor of 4 times even increases by a factor of 4 times even through the energy passing through through the energy passing through it only doubles. Explain qualitativelyit only doubles. Explain qualitatively

Soln:Soln: Doubling the width results in narrowing of Doubling the width results in narrowing of

the diffraction pattern As the width of the the diffraction pattern As the width of the central maximum is effectively cut in half, central maximum is effectively cut in half, then there is twice the energy in half the then there is twice the energy in half the space, producing four times the intensityspace, producing four times the intensity

Problem no 8Problem no 8 Calculate approximately the relative intensities of Calculate approximately the relative intensities of

the maxima in the single slit ,Fraunhofer Diffraction the maxima in the single slit ,Fraunhofer Diffraction patternpattern

Soln:Soln: The maxima lie app half way between the minima and are The maxima lie app half way between the minima and are

roughly given by roughly given by = (m== (m=½)½) where m=1,2,3….. where m=1,2,3….. IIөө = I = Im m {sin (m={sin (m=½)½)// (m= (m=½)½)}}22

IIөө / I / Im m == {1/ (m={1/ (m=½)½)}}22

==0.0450 for m=1,0.0450 for m=1, =0.0162 for m=2,=0.0162 for m=2, =0.0083 for m=3, =0.0083 for m=3, =0.0050 for m=4,=0.0050 for m=4, =0.0033 for m=5=0.0033 for m=5

Problem no 9Problem no 9 In a double slit experiment the distance D of In a double slit experiment the distance D of

the screen from the slits is 52cm the the screen from the slits is 52cm the wavelength is 480nm, the slit separation is wavelength is 480nm, the slit separation is 0.12mm and the slit width is 0.025mm0.12mm and the slit width is 0.025mm

A.what is the spacing between adjacent A.what is the spacing between adjacent fringes B.what is the distance from the fringes B.what is the distance from the cenetral maximum to the first minimum of cenetral maximum to the first minimum of the fringe envelopethe fringe envelope

Soln: Soln: y = y = λλD/d =(480x10 D/d =(480x10 -9) (-9) (52x10 52x10 -2-2)/(0.12x10)/(0.12x10-3-3) =2.1mm ) =2.1mm Angular separation of the first minimum is Angular separation of the first minimum is sinsinөө = =λλ/a = 0.0192 /a = 0.0192 Y = D tan Y = D tan өө = D sin = D sinөө =(52x10 =(52x10 -2-2)(0.0192) = 10mm)(0.0192) = 10mm There are about 9 fringes in the central peak of of the diffraction There are about 9 fringes in the central peak of of the diffraction

envelopeenvelope

Problem no 10Problem no 10 What requirements must be met for the What requirements must be met for the

central maximum of the envelope of the central maximum of the envelope of the double slit interference pattern to contain double slit interference pattern to contain exactly 11 fringes? How many fringes lie exactly 11 fringes? How many fringes lie between the first and the second minima of between the first and the second minima of the envelope?the envelope?

Soln:Soln: The required condition will be met if the 6 th min of the The required condition will be met if the 6 th min of the

interference factor (cosinterference factor (cos22ββ) coincide with the 1) coincide with the 1stst minimum of minimum of the diffraction factor (sinthe diffraction factor (sin//))22..The sixth minimum of the interference factor occur when The sixth minimum of the interference factor occur when d sind sinөө = 11 = 11λλ/2 or /2 or ββ = 11 = 11/2./2.

The first minimum in the diffraction term occurs for The first minimum in the diffraction term occurs for dsindsinөө = = λλ

Or Or = = and d/D = 11/2 or d=5.5 and d/D = 11/2 or d=5.5

Problem no 11Problem no 11 A. Design a double slit system in which the 4A. Design a double slit system in which the 4thth fringe fringe

not counting the central maximum is missing.not counting the central maximum is missing. B. what other fringes if any are also missing?B. what other fringes if any are also missing? Soln:Soln: A.A. d d sinsinөө =4 =4λλ gives the location of the 4 gives the location of the 4thth interference maximum. interference maximum. a a sinsinөө = =λλ, gives the location of the first diffraction minimum., gives the location of the first diffraction minimum. If d = 4a, there will be no 4If d = 4a, there will be no 4thth interference maximum. interference maximum.

B.B. d sind sinөөmimi = m = mmimiλλ gives the location of the m gives the location of the mthth interference maxima. interference maxima. d sind sinөөmdmd = m = mmdmdλλ gives the location of the m gives the location of the m thth diffraction minima diffraction minima D=4a hence if mD=4a hence if m i i =4m=4md d there will be a missing maxima there will be a missing maxima

Problem no 12Problem no 12 The wall of large room is covered with The wall of large room is covered with

acoustic tile in which small holes are drilled acoustic tile in which small holes are drilled 5.2mm from the center to the center. How 5.2mm from the center to the center. How far can a person be from such a tile and still far can a person be from such a tile and still distinguish individual holes assuming ideal distinguish individual holes assuming ideal condition? Assume the diameter of the pupil condition? Assume the diameter of the pupil of the observer’s eye to be 4.6mm and the of the observer’s eye to be 4.6mm and the wavelength to be 542nm .wavelength to be 542nm .

Sol:Sol: y/D = 1.22y/D = 1.22λλ/a (here a=4.6mm and y=5.2mm)/a (here a=4.6mm and y=5.2mm) D = 36.2mD = 36.2m

Problem no 13Problem no 13 The two head lights of an approaching automobile The two head lights of an approaching automobile

are 1.42 m apart . At what are 1.42 m apart . At what A) angular separation andA) angular separation and B) maximum distance will the eye resolve them?B) maximum distance will the eye resolve them? Assume a pupil diameter of 5 mm and a wavelength Assume a pupil diameter of 5 mm and a wavelength

of 562 nm. Also assume that the diffraction effects of 562 nm. Also assume that the diffraction effects alone limit the resolution.alone limit the resolution.

Solution:Solution: A. least angular separation required for the resolution is A. least angular separation required for the resolution is өөRR = sin = sin -1-1(1.22(1.22λλ/a) =1.37 x 10/a) =1.37 x 10-4-4 rad rad өөR R =y/D =1.42/D=1.37x10 =y/D =1.42/D=1.37x10 -4-4 rad. rad. D=1.04X 10D=1.04X 1044

Diffraction grating problemsDiffraction grating problems A certain grating has 10A certain grating has 1044 slits with a spacing d=2100 nm. It is slits with a spacing d=2100 nm. It is

illuminated with a light of wavelength 589 nm . illuminated with a light of wavelength 589 nm . Find Find A) The angular positions of all principal maxima observed and A) The angular positions of all principal maxima observed and B) the angular width of the largest order maximum.B) the angular width of the largest order maximum. Soln:Soln: A. d sinA. d sinөө = m = mλλ sinsinөө = m (589 x 10 = m (589 x 10 -9-9m)/(2100 x 10 m)/(2100 x 10 -9-9m)m) For m = 1, For m = 1, өө11 = 16.3 = 16.3 For m = 2, For m = 2, өө2 2 == 34.134.1 For m = 3, For m = 3, өө33= 57.3= 57.3 For m = 4, For m = 4, өө44= more than 90 degree hence 3.0 order is the = more than 90 degree hence 3.0 order is the

highesthighest B) for m=3, B) for m=3, өө = = λλ / Nd cos / Nd cosөө= 5.2 x 10= 5.2 x 10-5 -5 rad or 0.0030 degree rad or 0.0030 degree

Grating contd……Grating contd…… A diffraction grating has 10A diffraction grating has 1044 ruling uniformly ruling uniformly

spaced over 25 mm. It is illuminated spaced over 25 mm. It is illuminated normally using a sodium lamp containing two normally using a sodium lamp containing two wavelengths 589.0 and 589.59 nm.wavelengths 589.0 and 589.59 nm.

A. At what angle will the first order maximum A. At what angle will the first order maximum occur for the first of these wavelengths?occur for the first of these wavelengths?

B. what is the angular separation between B. what is the angular separation between the first order maxima for these lines. Will the first order maxima for these lines. Will this alter in other orders.this alter in other orders.

A. A. өө = sin = sin-1-1 m mλλ/d =13.6 degrees/d =13.6 degrees B. dB. dөө = m = mλλ/ d cos / d cos өө =2.4 x 10 =2.4 x 10-4-4 rads or 0.014 rads or 0.014

degrees.degrees. As the spectral separation increases with the order As the spectral separation increases with the order

no. this value increases with the order no.no. this value increases with the order no.

A diffraction grating has 1.2 x 10A diffraction grating has 1.2 x 1044 rulings uniformly spaced over a width w= rulings uniformly spaced over a width w= 2.5 cm and is illuminated normally using sodium light containing two 2.5 cm and is illuminated normally using sodium light containing two wavelengths 589 and 589.59nm.wavelengths 589 and 589.59nm.

A. at what angle does the first order maximum occur for the first of these A. at what angle does the first order maximum occur for the first of these wavelengthswavelengths

B. what is the angular separation between these two lines in the first and B. what is the angular separation between these two lines in the first and the second ordersthe second orders

C. how close in wavelength can two lines be in the first order and C. how close in wavelength can two lines be in the first order and өөλλ still be resolved by this gratingstill be resolved by this grating D. how many rulings can a grating have and just resolve the sodium D. how many rulings can a grating have and just resolve the sodium

doublet lines.doublet lines. soln:soln: A.A. өө = sin = sin-1 -1 (m(mλλ/d) = 16.4 degrees/d) = 16.4 degrees B.B. Dispersion D = Dispersion D = өө//λλ = m /(d cos = m /(d cosөө ) =5.0 x 10 ) =5.0 x 10-4-4 rad/nm rad/nm

өө = D x = D x λλ =2.95 x 10 =2.95 x 10-4-4 rads or 0.0169 degrees rads or 0.0169 degrees C.C. Resolving power = Nm = 1.2 x 10 Resolving power = Nm = 1.2 x 10 4 4

λλ = =λλ/R =0.049 nm hence can resolve the D lines./R =0.049 nm hence can resolve the D lines. D. R =D. R =λλ//λλ = 998. Hence no. of rulings needed is N=R/m =998/1=998 hence can = 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.easily resolve as it has 12 times no, of rulings in it.

Grating contd……Grating contd…… A grating has 200 ruling/mm and principal A grating has 200 ruling/mm and principal

maximum is noted at 28 degrees. What maximum is noted at 28 degrees. What are the possible wavelengths of the are the possible wavelengths of the incident visible lightincident visible light

Soln:Soln: λλ = (d sin = (d sinөө)/m = 2367 nm for m=1.)/m = 2367 nm for m=1. On trying for m =4 &5 we get in the visible On trying for m =4 &5 we get in the visible

range as 589nm and 469 nm and for m=6 range as 589nm and 469 nm and for m=6 and above it will be in the uv range.and above it will be in the uv range.

Grating contd……Grating contd…… For a grating the no. of rulings is 350/mm. A white For a grating the no. of rulings is 350/mm. A white

light falling normally on it produces spectrum 30 cm light falling normally on it produces spectrum 30 cm from it. If a 10 mm square hole is cut in the screen from it. If a 10 mm square hole is cut in the screen with its inner edge 50mm from the central maximum with its inner edge 50mm from the central maximum and parallel to it, what range of wavelengths passes and parallel to it, what range of wavelengths passes through the hole?through the hole?

Soln:Soln: Shortest wavelength passes through at an angle of Shortest wavelength passes through at an angle of өө11 = tan = tan

-1-1 (50mm/300mm)= 9.46 degree (50mm/300mm)= 9.46 degree λλ 11 ={ (1 x 10 ={ (1 x 10 -3-3)sin 9.46} /350 = 470 nm)sin 9.46} /350 = 470 nm The longest wavelength that can pass through an angle The longest wavelength that can pass through an angle өө22

= tan= tan--1(60mm/300mm) = 11.3 degree1(60mm/300mm) = 11.3 degree This corresponds to a wavelengthThis corresponds to a wavelength ΛΛ 22 = {(1x10 = {(1x10 -3 -3 )sin11.3} / 350 = 560 nm )sin11.3} / 350 = 560 nm

Grating contd……Grating contd……

A source containing a mixture of A source containing a mixture of hydrogen and deuterium atoms emit hydrogen and deuterium atoms emit light containing two closely spaced light containing two closely spaced red colors at 656.3 nm whose red colors at 656.3 nm whose separation is 0.180 nm. Find the separation is 0.180 nm. Find the minimum number of rulings needed minimum number of rulings needed in grating that can resolve these in grating that can resolve these lines in the first order.lines in the first order.

Solns;Solns;N = R/m = N = R/m = λλ/m/mλλ =365 =365

Grating contd……Grating contd……

A. How many rulings must a 4.15 cm wide A. How many rulings must a 4.15 cm wide diffraction grating have to resolve the diffraction grating have to resolve the wavelengths 415.496 nm and 415.487 nm in wavelengths 415.496 nm and 415.487 nm in the second order.the second order.

B. at what angle are the maxima foundB. at what angle are the maxima found Soln:Soln: N=R/m = N=R/m = λλ/m/mλλ = 23100 = 23100 D = w/N = …..D = w/N = ….. өө = sin = sin -1-1 m mλλ/d = 27.6 degrees/d = 27.6 degrees