exercise –1 part - i · 2020. 11. 24. · wave optics reg. & corp office : cg tower, a-46...

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EXERCISE –1 PART - I

Section (A) A-1. We know that res = 1 + 2 + 2 1 2 cos ()

(a) res = + 4 + 2 × 2 cos (0) = 9 I.

(b) res = + 4 + 2 × 2 cos 2

= 5 I.

(c) res = + 4 + 2 × 2 cos (180) = I. A-2. we know that equation of light wave E = E0 sin (t – kx )

speed of wave v = k

by comparing the given equation v = 3 × 104 × 5 × 10–7 v = 15 × 107 m/s

Here 14x ox 7

yE (y,t) E sin2 3 10 t

5 10

= vk

Now =C

v =

7

7

30 10 m / s

15 10 m/ s

= 2.

Section (B)

B-1. We know = D

d

(a) if D is increased increases. so the width of interference fringes increases. (b) decreases decrease again. (c) d is increased decreases. If the source slit is moved nearer. no effects on , D or d, so remains same. (d) By slightly increasing the width of the slits ,we are only increasing the intensity of incident

beam.Again no change in , D, d. so unchanged but sharpness of the fringe increase. B-2. As D >>d & << d.

Hence we can use = D

d

so distance between 5th bright fringe and 3rd dark fringe = 5 – (2 + /2)

= 5

2 =

–7

3

5 6.5 10 1

2 10

= 1.625 mm

B-3. AS D

d

= 0.4 mm

by changing ; becomes

’ =3

4 × 0.4 = 0.3 mm.

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B-4. As << d; we can use

=D

d

so

D d

=

10

3

6000 10

3 10

= 2 × 10–4 radian

=180

× 2 × 10–4 degree = 0.011º

Section (C) C-1. As << d

1 = 1 D

d

2 = 2 D

d

So, 2 – 1 = 2 1– D

d

=

9

3

700 – 580 10 1.5

0.20 10

= 0.9 mm.

Section (D) D-1. (a) By inserting the strips, does not change.

= D

d

=

9

2

480 10 1 m

0.12 10

= 4.0 × 10–4 m.

(b) Lets try to find out the point at which path difference is zero. (central maxima).

So at P; path difference due to extra travelling (i.e. dsin) must be equal to the optical path difference. at centre point path difference = (2 – 1) t = 83.33

so as we move up path difference will decrease so at height 3

we will get maxima and at a distance

2

3

on lower side maxima will formed.

D-2. Clearly, optical path difference

= ( – 1) t = 4

t =

4( – 1)

.

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Section (E) E-1.

clearly, at O, no difference of path due to inside motion of rays. only path difference = S1 Q = d cos (90–) = d sin.

= d × 2d

=

2

Net phase difference

so = 2

2

± ( – 1)t.

2

= ± ( – 1) 2( 1)

2

= 2, 0

constructive interference and hence Ires = maximum

Section (F) F-1. The path difference is 2t. Now for destructive interface it can be

2t = 2

or

3

2

or

5

2

and so on .........

= 4t

, 3

4t

,

5

4t

.....

= 9

6

580 10

4 0.3 10

, 9

6

3 580 10

4 0.3 10

................

= 0.4833 , 3 × 0.4833.............. so only = 3 × 0.4833 = 1.45 is the answer, {1.3 < < 1.5} F-2. Clearly for weak transmition, it should be destructive interference.

2t = 2

{for smallest thickness}

t = 4

= 9560 10

4 1.4

= 100 nm

Section (G) G 1. (i) For first maximum

sinm =

1m

2

a

sin1 = 3

2a

or a = 1

3

2sin

= 73 6 10

2 sin30º

= 1.8 m

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(ii) For first minimum

sinm = m

a

or sin 1 =a

a = 1sin

a = 76 10

sin30º

= 1.2 m

PART - II Section (A) A-1. we know I A2.

2

1 12

2 2

A

A

4

1 1

2

A

A A1 : A2 = 2 : 1

A-2. Imax = 21 2 = 24 = 9I.

Imin = 21 2– = 24 – = I.

Section (B) B-1. Contrast indicates the ratio of maximum possible intensity on screen to the minimum possible intensity.

As max

min

=

2

1 2

2

1 2–

so it only depends on the source intensity.

B-2 w = D

d

since v = f since vacuum is made, increased fringe width increases

B-3. we know that =D

d

& yellow > blue.

as decreases, so also decreases.

B-4. As << d; we can we = D

d

we get = 9

3

500 10 1

10

= 0.5 mm.

As is not very small; hence it might so happen that till 1000th maxima, we no longer can apply y’ = 1000 × . Lets see if we can apply: At 1000th maxima. Path difference is 1000 .

1000 = d sin = 2 2

d y

D y

(5 × 10–4)2 = 3 2 2

2 2

(10 m) y

D y

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0.25 D2 = y2 (1 – 0.25) y =

1

20.25D

0.75

y = D

3 = 0.577 m.

As 0.577 m. and 0.5 m. are quite distant, so we could not use y’ = 1000 for such a high maxima. B-5. Lets look at the screen.

As we know that 75% intensity will correspond to a point where intensity is 3 0. {Imax = 4 I0}

I = I0 + I0 + I I0 02 cos

3I0 = 2I0 (1 + cos )

cos () =

= 3

, 2 –

3

, 2 +

3

,...........

p = 6

, –

6

, +

6

,...........

p = yd

D

3I0

D

yd

D =

6

y =

D

d ×

6

,........

y = 6

, –

6

, +

6

ymin. = 3

y =3

×

D

d =

–10

–3

6000 10 1

3 10

= 0.2 mm

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B-6. Lets take any general point S on the line AB.

Clearly: for any position of S on line AB; we have for PQS: PQ + QS > PS {in any triangle sum of 2 sides is more then the third side} PS – QS < 3. As PS – QS represents the path difference at any point on AB it can never be more than 3. Now

minimas occur at.

3 5

, ,2 2 2

only.

so 3 minimas below R (mid point of AB) and 3 also above R.

B-7._ 2R 0I I cos

2

2x

3

2RI cos

6

0

I 3

I 4

Section (C) C-1.

we know that P will be the central maxima (at which path difference is zero)

Now OP = d d d

–2 3 6

C-2. Fourth maxima will be at y = 4.

y = 4 D

d

as Green > blue. Green > blue XGreen > Xblue

Also get X(blue) 4360

X(green) 5460

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Section (D) D-1.

Clearly the central maxima at P(initially) shifts to P’ where PP’ = 5 mm. So now, path difference at P’ must be zero. d sin = ( – 1)t d tan = ( – 1)t

= 1 + d.(PP')

Dt ; get = 1.2

D-2. As we know, at the point of 75% intensity

cos = 1

2 {Refer Q B–4, Exercise –I Objective}

2

× P = 3

,5

3

,7

3

,11

3

,13

3

(µ – 1) t = 5 7 11 13

, , , ,6 6 6 6 6

t = 5 7 11 13

, , , ,6( – 1) 6( – 1) 6( – 1) 6( – 1) 6( – 1)

= 0.2 m; 1 m, 1.4 m, 2.6 m....... Hence only 1.6 m is not possible. Section (E)

E-1. (t1 – t2) (n – 1) = 2

[(t1 + t2)] (n – 1) = 2 3t1 = 5t2 – 1

21

21

tt

)tt(

= 4

1 4t1 – 4t2 = t1t2

= 3t1 = 5t2

= 1

2

t

t =

5

3

x = 15. E-2. From the figure shown below it is clear that the distance of central maxima from O is D.

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Section (F) F-1. For strong reflection.

2t = 3 5

, ,2 2 2

......

= 4t, 4 t 4 t 4 t

, ,3 5 7

......

3000 nm. 1000 nm, 600 nm, 430 nm, 333 nm.

Section (G)

G-2. 1.22λ

dθa

G-3. When unpolarized light passes through a Polaroid, its intensity becomes 50%.

G-4. a

Rv Rv

G-5. By Brewester law reflected light will be a plane polarized light with vibrations Perpendicular to the plane

of incidence

G-6. D

mm1

a22.1

a = 3mm, = 500nm

G-7. a

=

D

R

and x-Ray < visible Rx-Ray< 0.1

PART - III 1. Initially at a distance x from central maxima on screen is

= 0 + 40 + 0 02 4 cos2 x

, where = D

d

max = 90 and min = 0

(A) At points where intensity is1

9 th of maximum intensity, minima is formed

Distance between such points is , 2, 3, 4, .....

(B) At points where intensity is3

9 th of maximum intensity, cos

2 x

= – 1

2or x =

3

.

Distance between such points is2

,3 3

, , +

3

, + , 2, .....

(C) cos 2 x

= 0 or x = 4

.

Distance between such points is , , +2

3

, 2, .....

(D) cos 2 x

=1

2 or x =

6

.

Distance between such points is , 2

, + , +

2

, 2, .....

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2. The y-coordinate of point P = D

2d

= half the fringe width. Hence initial intensity at P is zero.

(A) As d is doubled, fringe width becomes half. As a result first order bright is formed at P and optical path difference between interfering waves at P will increase.

(B) As D is doubled, fringe width doubles, hence intensity at P cannot be zero. Also optical path difference between interfering waves at P will decrease.

(C) Decreasing width of slit will reduce energy from S1. But as D is doubled intensity at P is changed from zero to non zero value. Hence all options are same as in case B.

(D) As set up is submerged in water of refractive index4

3 , fringe width reduces, optical path difference

will become4

3 rd its initial value. Hence dark cannot be again obtained at P. Therefore intensity will

increase P.

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EXERCISE–2

PART - I

1. max min

max min

–

=

2 2

1 2 1 2

2 2

1 2 2 2

– –

–

= 1

1

×

2 2

2 2

1 1

2 2

2 2

1 1

1 – 1–

1 1–

= 2 2

2 2

(1 2) – (1– 2)

(1 2) (1– 2)

= 8 4

10 5

2. Obviously; for = 1, O will be a maxima. As increases, the intensity will decrease and hence option

(C). 3. Let us say, nth minima of 400 nm coincides with mth minima of 600 nm.

1

n2

D400

d = =

1m

2

. D

600d

400 n = 600m + 100.

n = 6m 1

4

= (some integer or non-integer) + 0.25

Hence n can never be an integer. So no minima of 600 nm coincides with any minima of 400 nm. 4. Notice that d is not very small than D; so we can not use p = d sin.

S1P – S2P = P =2

{ first minima}

2 25 12 – (12) = 2

get = 2 cm

5. At B; P = 4 (maxima) At x = ; P = 0 (maxima) Hence in between; the point at which path difference is either , or 2 or 3 they will be maximas.

Hence 3 maximas.

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6. (c)

s1 s2

d

x dcos n

d 2

n

cos2

n 1

1

cos2

60 7. When light passes through a medium of refractive index , the optical path it travels is (t). Therefore, before reaching O light through S1 travels (l + b) distance while that through S2 travels a

distance (l + b)

i.e. : path difference = (l + b) – (l + b) = ( – 1) l. For a small element 'dx' path difference x = [(1 + ax) – 1] dx = ax dx For the whole length ;

x = 2

O

aax dx

2

For a minima to be at 'O'.

x = (2n + 1) 2

i.e. : 2a

2

= (2n + 1)

2

.

For minimum 'a'; n = 0

2a

2 2

a =

2

Ans.

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8. The 2 sources are.

As O is a maxima, Hence OP = .

d .D

2 (3d)

; get

23d

2D

9. Point O is a minima Hence the first maxima will be at y =2

from O.

y = D

2(2d)

=

–9

–3

600 10 1

4 1 10

= 0.15 mm.

10. A B

C D

for first dark fringe at P

A

B

C

D

P

also C ~ B A – D = 2 11._ a sin 30° = a = 2

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PART – II 1.

Say ‘n’ fringes are present in the region shown by ‘y’

y = n = n. D

d

y 0.06 n

tan(0.06º )D 180 d

n =310

0.06180

=

3

> 1.

Hence; only one maxima above and below point O. So total 3 bright spots will be present (including point ‘O’ i.e. the central maxima).

2.

(i)

Q is equidistant from s1 and s3 Resultant amplitude due to light coming from S1 and S3 is AR = 2A Resultatnt amplitude at Q

AR = 3A

23A = (2A)2 + A2 + 2 2A × A cos

3A2 = 5A2 + 4A2 cos

–1

2 = cos, = 2

3

,

4

3

x = 2 2D d – D

= D 2

2

d1 –1

2D

=

2d

2D

2d

2D × 2

=

2

3

= 23d

2D

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(ii)

Intensity for point 'P'

for S1 and S2 at P = 2d 2 2

. 1202D 3

for S1 and S3 at P = 22.d 2 8

. 4802D 3

Again Area = 2 2 2A (2A) 2A cos120 = 3 A

IP = 3I

120°

A

2A=A+A

3. Where Z = D

2d

=

2

OS4 =

4

as shown.

If intensity at 'p' is then intensity of light at S3 and S4 is /4 & /4 Path difference S4P – S3P = 0 So, intensity of slits S1 and S2

at S4 S4

p = yd

D =

d z

D 2

= 4

= 2

.4

= 2

4

= R = 1 + 1 + 1 12 cos

2

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1 = 2 = 8

intensity of S1 and S2 .

If z = D4.

d

= 4

p = yd

D= z d

.2 D

= 4 D 1 d

d 2 D

= 2

= 2

.2d = 4

3 = 4 = 2

cos48 8 8 2

at 'p' p = 2 . cos0º2 2 2 2

p = 2. 4. The shown diagram represents the reflection and refraction intensity.

Hence the 2 waves interfering, 1 and 2, have intensities 0

4

and 09

64

.

min

max

=

2

0 0

2

0 0

9–

4 64

9

4 64

=

2

2

1 3–

2 8

1 3

2 8

= 1

49

5. 1 = 1.4, 2 = 1.7 and let t be the thickness of each glass plates. Path difference at O, due to insertion of glass plates will be -

1

2

S1

S2

O

6 Minimath

5 Minimath

x = ( 2 – 1 ) t = ( 1.7 – 1.4 ) t = 0.3 t .......(1) Now since 5th maxima ( earlier) lies below O and 6th minima lies above O. This path difference should lie between 5 and 5 + /2 So let x = 5 + .......(2) where < / 2

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Due to the path difference x, the phase difference at O will be

= 2

x = 2

( 5 + ) = 10 +2

. ) .......(3)

Intensity at O is given 3

4 Imax and since

I () = Imax cos2

2

3

4 Imax = Imax cos2

2

or 3

4 = cos2

2

.........(4)

From equations (3) and (4), we find that = / 6 i.e. x = 5 + / 6 = (31 / 6) = 0.3 t

t = 31

6(0.3)

=

10(31)(5400 10 )

1.8

m

or t = 9.3 × 10-6 m = 9.3 m Ans. (14)

PART - III

1.* =

2

1 2

2

1 2

9

1–

by checking the options : 1 = 4 unit. 2 = 1 unit.

and 1 1

2 2

A

A

= 2.

2.* As d << D, path difference = d sin (at 0) = 1mm × sin 30º = 0.5 mm if it is a maxima 10–3 × 0.5 = (5000 × 10–10 )m × (n) n must be integer. get n = 1000. Hence O is a maxima of intensity 4I0 Now

Now path difference at Q = d sin only QS1 QS2. d sin = 1 × 1/2 = 0.5 mm = integer multiple of . Hence maxima.

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3.* If it is performed with white light. the central point will have maxima of all the colours, hence it will look white.

D

d

; as v is minimum

so first maxima after white as will be that of violet. So there will be no dark fringe as it is not possible be have minimas of all the colors at the same point.

4.* for = 30º; P = d sin 30º = (150 m) × 1

2 = 75 m.

= 75 × 2

= 8

6

275

3 10

10

= 2

Ires = I + I + 2I cos () = 2I

where I = 0

4

{ maximum intensity = I0 = 4I} Ires = 0

2

= 0º will automatically correspond to central maxima.

= 90º will correspond to path diff = d. = 150 m = 8

6

150 2

3 10

10

=

minima Hence intensity will be 0 at = 90º. Alternate

= 30º P = d sin 30º = (150 m) × 1

2 = 75 m.

= 75 × 2

= 8

6

275

3 10

10

= 2

Ires = I + I + 2I cos () = 2I 5*.

Clearly at Q, path difference = d sin

b sin btan b.y

d =

2b

2d

Now whenever 2b

2d will be odd multiple of

2

, those ’s will be having minima at point Q.

2b

2d =

2

, 3

2

,

5

2

....

= 2b

d ,

2b

3d,

2b

5d .....

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PART - IV 1. Wave fronts are spherical in shape of radius ct. Hence (D). 2.

The wave fronts are always perpendicular to the light rays. Hence, (C). 3. Using snell's law ;

0sin(45 ) 2

sinr 1 sinr =

1

2 r = 300

Hence, (B) is correct. Note : The shown lines are wavefronts and not rays. 4. After reflection by mirror the parallel rays concentrate at the focus.

Hence the plane wave front becomes spherical concentrated at the focus. Hence, (A).

5. In ABC ; sin (i) = 2

d In xyz ;

sin (r) = 1

d

sin i

sin r= 2 = .

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6. If phase difference at point P is zero then n1 d sin = n2 d sin'

dsin

dsin

yP

' = 37°

and as tan' = y

D y = –

3

4m

It is negative because upper path in medium n2 is longer than lower path in the same medium. 7. Path lengths in medium 2 are equal for point O. Therefore path difference = d sin

1n 0.3 mm,

2n

4(0.3)

30.36 mm

109

4dsin

3p10

9

,

=2p

2 1 4 / 3

14 / 3 2 10 / 9

0.310 / 9

= 10

3

= 0 + 0 + 20 4

cos 23

= 0

8. As we go up from point O, path difference will increase. At O, phase difference is 3 +3

and when it

becomes 4, there will be maximum. Extra path difference created in medium 2 must lead to 2

3

phase

difference.

a

2

. d sin1. n2 = 2

3

Using values sin1 = 3

25 tan 1 =

3

616 =

y

D

y = 300

2 154cm =

150

154cm

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EXERCISE –3 PART - I

1.

Wavefronts (Normal to rays) are parallel. So rays should also be parallel. 2. c and d are at same wavefronts So, C = d and e = f

d – f = c – f D is incorrect d – c = 0 f – e = 0 So both should be equal. 3. Ray is bending towards normal. So, medium 1 should be rarer and medium 2 should be denser. So, V1 > V2 . 4. If d = , then maximum path difference will be less than . So there will be only central maximum on the

screen. If < d < 2, then the maximum path difference will be less than 2. So there will be two more maximum on screen

corresponding to path difference x = So (A) and (B) are correct. Intensity of dark fringe becomes zero when intensities of two slits are equal. Initial intensity at both the

slits are unequal so there will some brightness at dark fringe. Hence when intensity of both slits is made same the intensity at dark on screen shall decrease to zero. So both (C) and (D) are false.

Ans. (A) and (B)

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5. (A) (P1) = 1 + 2 + 2 1 2 cos 2

4

= 0 + 0 + 20 . 1

2= (2 + 2 ) 0

(P2) = 1 + 2 + 20 . 1

2 = 2 1 2 . cos

2

3

= 0 + 0 + 20 . 1

2

= 0

(P1) > (P2)

(B) (P0) = ( – 1)t . 2

= 2

.4 2

(P1) = 2

( 1)t . 04

(P2) = 2

( 1)t .3 6

(P0) = 1 + 2 + 2 1 2 . cos (P0)

= 0 + 0 + 20 . cos 2

=20

(P0) = 40

(P2) = 0 + 0 + 2 0 cos 6

= (2 + 3 )0

(C). (P0) = ( – 1)t . 2

= 2

.2

(P1) = 2 2

( 1)t.4 2

(P2) = 2

( 1)t .3 6

(P0) = 2 2

.2 3 6 3

(D) (P0) = 3 2 3

4 2

(P1) = 0 (P2) > 0, (P0) > 0

6. = D

d

VIBGYOR

increase R > G > B So R > G > B

7. For half of maximum intensity 20= 0 + 0 + 20 cos

(Phase difference) =3 5

, , .........2 2 2

Path difference is 3 5 2n 1

, , .........4 4 4 4

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8. 2d sin =

d = 2sin

differntiate

(d) = 2

(cosec)

(d) = (– cosec cot )2

(d) = 2

– cos

2sin

as = increases , 2

cos

2sin

decreases

Alternate solution

d2sin

nd n n2 nsin

(d)

d

= 0 – 0 –

1cos ( )

sin

Fractional error |+(d)| = |cot | Absoulute error d = (dcot)

d cos

2sin sin

2

cosd

sin

9. =D

d

2 > 1 so 2 > 1

No of fringes in a given width (m) =y

m2 < m1

3rd maximum of 2 = 23 D 1800 D

d d

5th minimum of 1 = 19 D 1800 D

2d d

So, 3rd maxima of 2 will meet with 5th minimum of 1

Angular sepration = d

Angular seperation for 1 will be lesser

= D

d

2 > 1 2 > 1

10. For constructive interference x = m

4

32 2d x – 2 2d x = m

1

32 2d x = m

x2 = 9m2 2 – d2 p = 3

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11*. from theory fringes will be semi circular

and 2

11000

d

at 0 x = 2

1000

so at 0 it will be dark 12.

P1

P2 S2 S1

d

dsin

x=dsin

= 600 nm

at P1 x = 0

at P2 x = 1.8mm = n

No. maximum will be = n = x

= nm600

mm8.1 = 3000

at P2 x = 3000 hence bright fringe will be formed.

at P2 3000th maxima is formed. for 'D' option

x = dsin

dx = dcos.d

R = dcos.Rd

Rd =

cosd

R

as we move from P1 to P2

cos Rd

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PART - II

1. Phase difference = 2

× path difference

i.e = 2

× 6

=

3

As I = Imax cos2 2

max

= cos2 2

0

= cos2 6

0

= 3

4

3. Beam does not converge or diverge so shape of wave front remain planar. 4. = 0 + 2 ()

r

as it is given that intensity of beam is decreasing with increasing radius, and as decreases also

decreases.

Now by V = C

speed of light is minimum at axis of cylinder. 5. x1 = 0 = 0º 1 = 0 + 0 + 20 cos0º = 40

x2 = 4

= 2

.4

= 2

2 = 0 + 0 + 20 cos 2

= 20

1

2

= 0

0

4

2

=2

1 .

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6. For coherent sources : 1 = 40 For incoherent sources 2 = 20

0

2

=2

1 .

7. The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a

polaroid which is rotated.

8. S1 : When light reflects from denser med. (Glass) a phase dift of is generated. S2 : Centre maxima or minima depends on thickness of the lens.

9. It will be concentric circles

10.

IA cos230° = IBcos2 60° A 3

4= B .

1

4

A

B

1

3

11.

Ray 2 will travel faster than 1, so beam will bend upward

Wave Optics




12.

Resolving angle of necked eye is given by :

1.22D

–9

–2 –2

y 1.22 500 10

25 10 0.25 2 10

y = 30 × 10–6 m = 30 m

13. b radius of spot.

b = a + a

L

geometrical spread + spread due to diffraction

da

db = 0

1 – La2

= 0 a2 = L a = L

bmin. =

L

LL

bmin. = L2

bmin. = L4

14. 4

9

1105

5.110650

d

D

= 650 × 0.3 × 10–5 = 1.95 mm

4

9

2105

5.110520

= 520 × 0.3 × 10–5 = 1.56mm 41 = 52 = 7.8 mm

Wave Optics




15.

I

A C B

I/2

I/2 cos2

B

I/2 cos2 cos2

I

2cos2 =

I

8

cos2 = 1

4

cos = 1

2 = 45°

16.

30° 30°

Central maxima

Slit

Semi-angular width = 30° asin =

asin30° =

Fringe width D

d

=

asin30 D

d

6

2

110 0.5

210d

d = 5510

2 = 25m

Wave Optics




SOLUTIONS OF HIGH LEVEL PROBLEMS (HLP) SUBJECTIVE QUESTIONS

1. Values of , etc. are given because there were 2 more parts asked in actual JEE paper based on value of ‘b’ and . But for finding the intensity, we don’t need any of this.

The paths of 2 rays is as shown. Clearly the path difference between them is. PQ – RS = (PS) cos(90–r) – .d tan = (PS) sin r – d tan. Now ; by snells. law.

sin = 1. sin r. and PS = d

cos

we get path difference = d sin

cos

– d tan = 0 .

Hence the intensity is maximum at 9I.

2. As << d, = D

d

can be used

For minima

(a) y = 1

n2

.= 1

n2

D

d

= –3 –3 61.5 10 1 10 1.5 10

11 n1 n22

n = 1, = 2

3 × 1.5 × 10–6 = 1000 nm

n = 2, =2

5 × 1.5 × 10–6 = 600 nm

n = 3, = 2

7 × 1.5 × 10–6 = 428 nm

Putting integral values of ‘n’ n = 1 ; = 1000 nm n = 2; = 600 nm. n = 3; = 428 nm So only = 428 nm and = 600 nm, will have minima at the hole. Hence they will be absent in the light

coming out. (b) 1.5 mm = n .

1.5 mm =D

n d

= 3 3

2

1.5 10 1 10

n 100 10

for n = 1. = 1500 nm for n = 2. = 750 nm. for n = 3. = 500 nm. Hence only = 500 nm will have maximum intensity.

Wave Optics




3. (a) Clearly 3rd bright fringe will be at y = 3.

= 3 D

d

=

–10

–3

3 (6500 10 ) 1.2

2 10

= 0.117 cm. = 1.17 mm.

(b) Say mth bright fringe of 6500 Å coincides with nth bright fringe of 5200 Å . n1 = m2

n

m =

5

4

Hence, for least distance, 5th bright fringe of 5200 Å coincides with 4th bright fringe of 6500 Å.

y’ = 4 1 =–10

–3

4 (6500 10 ) 1.2

2 10

= 0.156 cm = 1.56 mm

4.

at O; path difference = dsin where is the angle as shown: (as >> d)

so P= d sin d tan d.d/ 2

=

2d

2

= 22 d

2

=

2d

.

(a) Now if a liquid is filled as shown.

path difference = (SS2 – SS1) + S2 P – S1 P

= 2d

2 +

2d

2D

phase difference = 2d 2

2

+ 2d 2

2D ( / )

= 2d 1

D

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(b) (i) if liquid is filled on the left side.

=2d 2

2 /

+ 2d 2

2D

= 2d

1

D

(ii) Clearly it is asking about /2 { << d}.

Dmin = D

2 2d

5.

After the thin plate is placed between one of the slits and the screen, an additional path difference [P =

( – 1) t ] is created between the rays emerging out of S1 and S2. Thus S1O – S2O = (– 1) t = 0.75 × 10–6 m

( – 1)t

=

3

2 ( – 1)t =

3

2

Point O would be an interference minimum. Thus intensity at O is zero.

S2P – S2P = mdy

D

S2P – S1P = mdy

D– ( – 1)t

For central maxima, the path difference between the interfering beams is zero. i.e. S2P – S1P = 0

mdy

D = ( – 1)t

ym = D

d ( – 1)t =

4

1

5 10 ×0.75 × 10–6 = 1.5 mm

Thus, the effect of putting a thin film in front of one of the slits is to shift the central maximum towards

that slit.

6. Clearly initial = i

D

d

0.25 × 10–3 = d

× 1 m

d

= 2.5 × 10–4 .............(1)

Wave Optics




Afterwards :

i

2

3 =

D

(d d)

d d

= 42 2.5 10

3 1

.............(2)

Dividing (1) and (2)

d d

d

=

3

2 d = 2 (d) = 2.4 mm.

& = 2.4 × 2.5 × 10–7 m = 600 nm.

Now Now P becomes central maxima. for point P : d sin = ( – 1) t

d. 20

D

tan ( – 1)t

d = ( – 1)t

d 20 D

D d

= (1.5 – 1)t

t = 20

0.5

=

–920 600 10

0.5

= 24 m.

7. (a) For maximum thickness of B. the central maxima is just above O.

Phase difference at O must be = 2

3

{ cos = –

1

2 for 3I intensity}.

Path difference at O = 2

3 2

=

3

d sin 30° + (A – 1) × tA – (B – 1) tB = /3 get tB = 120 m. (b) Now tB = 120 m.

= D

d

=

–10

–3

6000 10 1

0.1 10

= 6 mm.

Imax = 24 = 9 I

Imin = 24 – = I

(c) P path difference at 'O' P = dsin + (A –1) tA – (B – 1) tB = 0.1 sin30º + (1.5 – 1) × 20.4 – (1.5 – 1) × 120 P = 2 × 10–7 m

y d

D= P

y = D

Pd =

–3

1

0.1 10 × 2 × 10–7 = y = 2 mm

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= D

d

= 6mm

distance between maxima and minima = 2

= 3mm

distance of nearest minima from 'O' = 6

= 1mm

(d) So phase difference at P1

=yd

D ×

2

= –3 –448 10 10

1

×

–10

2

6000 10

= 16 Intensity at 'P'1

I1 = 4I + I + 22 4 cos (16)

I1 = 9 I Intensity at point 'P2' phase difference at 'P2'

= 2

× P

P = yd

D =

52mm 0.1 mm

1 m

= –10

2

6000 10

× 52mm 0.1 mm

1 m

= 523

= 16 + 4/3

I2 = 4I + I + 22 4 cos (16+4/3)

I2 = 4I + I + 22 4 cos (4/3)

= 4I + I – 2 × 2I × 1

2

I2 = 3 I

Wave Optics




8. Lets find out the radius of curvature of equi. convex lens.

1 1 1

( – 1) –f R –R

1 3 2

– 110 2 R

R = 10 cm.

Now

for lens :

1 1 1

–V –20 10

1 1

V 20

for surface of tube (of R = 10 cm.) (R = 10 cm.)

2 1 2 1––

V u R

4 / 3 1 4 / 3 – 1–

V 20 –10

V = + 80 cm.

(b) Now for mirrors. As the object for the mirrors is at 20 cm so the image will be at 20 cm only u = – 2 f v = – 2f also.

magnification = m = 0

y –v

y u

–y –20

–(1 mm) –20

yI = + (1 mm)

so the final images are like.

so the distance between the images is 4 mm. (c) Now, these I2 and I4 behave as the 2 sources for fringe pattern.

= D

d

=

vD

fd =

(c / ) D

fd

= 8

15 –3

3 10 0.84 3

10 (4 10 )3 4

= 60 m.

Wave Optics




9. Given – = 600 mm = 6 × 10-7 (in medium) d = 0.45 mm = 0.45 × 10-3 m D = 1.5 m Thickness of glass sheet, t = 10.4 m = 10.4 × 10-6 m Refractive index of medium sheet,

m = 4/3

and refractive index of glass sheet, g = 1.5

(a) Let central maximum is obtained at a distance y below point O. Then

x1 = S

1P – S

2 P =

yd

D

Path difference due to glass sheet

x2 = g

m

1

t

Net path difference will be zero when –

x1 = x

2 or

yd

D = g

m

1

t y = g

m

1

t

D

d

Substituting the values, we have

y = 1.5

14 / 3

6

3

(10.4 10 )(1.5)

0.45 10

y = 4.33 × 10-3 m Ans.

(b) At O, x1 = O and x

2= g

m

1

t

Net path difference,x = x2

Corresponding phase difference, or simply = 2

. x

Substituting the values, we have

= 7

2

6 10

1.51

4 / 3

( 10.4 × 10-6 ) =13

3

Now I () = Imax

cos2 2

I = Imax

cos2 13

6

I = 3

4I

max Ans.

(c) At O : path difference is x = x2 = g

m

1

t

For maximum intensity at O x = n ( Here n = 1, 2, 3......)

= x x x

, ,1 2 3

............. and so on

Wave Optics




= 1.5

14 / 3

( 10.4 × 106 m) = 1.5

14 / 3

( 10.4 × 103 nm )

= 1300 nm

Maximum intensity will be corresponding to

= 1300 nm, 1300

2nm,

1300

3nm,

1300

4nm.........

= 1300 nm, 650nm, 433nm.33nm, 325nm....... The wavelengths in the range 400 to 700 nm are 650 nm and 433,33 nm 10. Incident ray AB is partly reflected as ray 1 from the upper surface and partly reflected as ray 2 from the

lower surface of the layer of thickness t and refractive index 1 = 1.8 as shown in figure. Path difference

between the two rays would by x = 2 1t = 2 (1.8) t = 3.6 t

Ray 1 is reflected from a denser medium, therefore, it undergoes a phase change of , where as the

ray 2 gets reflected from a rarer medium, therefore, there is no change in phase ray 2. Hence phase difference between rays 1 and 2 would be = . Therefore, condition of constructive interference will be –

x = 1

n2

, where n = 0,1,2,3........

or 3.6 t = 1

n2

Least value of t is corresponding to n = 1 or

tmin

= 2 3.6

or tmin

= 648

7.2nm

or tmin

= 90 nm

(i) For a wave (whether it is sound or electromagnetic), a medium is denser or rarer is decided from

the speed of wave in that medium. In denser medium speed of wave is less. For example, water is rarer for sound, while denser for light compared to air because speed of sound in water is more than is air while speed of light is less.

(ii) In transmission / refraction, no phase change takes place. In reflection, there is a change of phase of when it is reflected by a denser medium and phase change is zero if it is reflected by a rarer medium.

(iii) If two waves in phase interfere having a path difference of x ; then condition of maximum intensity

would be x = n : n = 0,1,2....... But if two waves, which are already out of phase (a phase difference of .) interfere with path difference

x, then condition of maximum intensity will be

x = 1

n2

: n = 0,1,2,......

exercise –1 part - i · 2020. 11. 24. · wave optics reg. & corp office : cg tower, a-46...

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