804 + Interview Questions in OPTICS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Red light of wavelength `6500 Å` from a distant source falls on a slit `0.50 mm` wide. Calculate the distance between first two dark bands on each side of central bright band in the diffraction pattern observed on a screen placed `1.8 m` from the slit.
Answer» Here, `lambda = 6500 Å = 6.5 xx 10^(-7)m, a = 0.50 mm = 5 xx 10^(-4)m, D = 1.8 m` Distance between first two dark bands on each side of central maximum is the width of central maximum i.e., `2x = (2 lambda D)/(a) = (2 xx 6.5 xx 10^(-7) xx 1.8)/(5 xx 10^(-4)) = 4.68 xx 10^(-3)m`

Discussion

2.	The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the distance between two fringe from the centre is
Answer» Total intensity at any point is given by `I = kR^(2) = k (a^(2) + b^(2) + 2 ab cos phi)` when `b = a, I = k (a^(2) + a^(2) + 2 a^(2) cos phi)` `= 2 ka^(2)(1 + cos phi)` At the centre of a bright fringe, `phi = 0^(@)` `:. I_(1) = 2 ka^(2)(1 + cos 0^(@)) = 4 ka^(2)` Distance between two fringes `= beta`, which is proportional to wavelength `(lambda)` Now `(lambda)/(4)` corresponds to a phase diff. `= (2pi)/(4) = (pi)/(2)` `:. I_(2) = 2 ka^(2) (1 + cos pi//2) = 2 ka^(2)` `:. (I_(1))/(I_(2)) = (4 ka^(2))/(2 ka^(2)) = (2)/(1)`

Discussion

3.	The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the distance between two fringe from the centre isA. `2`B. `1//2`C. `4`D. `16`
Answer» Two waves of a single source having an amplitude `A` interfere. The resulting amplitude `A_(r)^(2)=A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)Cosdelta` where `A_(1)=A_(2)=A` and `delta=` phase difference between the waves `rArr I_(r)=I_(1)+I_(2)+2sqrt(I_(1)I_(2))Cosdelta` When the maxima occurs at the center, `delta=0` `rArr I_(r_(1))=4I`.....(`1`) Since the phase difference between, two successive frings is `2pi`, the phase difference between two points separated by a distance equal to one quarter of the distance between the two, successive frings is equal to `delta=(2pi)((1)/(4))=(pi)/(2)radian` `rArrI_(r_(2))=4Icos^(2)((pi//2)/(2))=2I`....(`2`) Using Eqs. (`1`) and (`2`), `(I_(r_(1)))/(I_(r_(2)))=(4I)/(2I)=2`

Discussion

4.	Among the two interfering monochromatic sources A and B, A is ahead of B in phase by `66^@`. If the observation be taken from point P, such that `PB-PA=lambda//4`. Then the phase difference between the waves from A and B reaching P isA. `156^(@)`B. `140^(@)`C. `136^(@)`D. `126^(@)`
Answer» Total phase difference `=` initial phase difference `+` phase difference due to path `=66^(@)+(360^(@))/(lambda)xxDeltax` `=66^(@)+(360^(@))/(lambda)xx(lambda)/(4)=66^(@)+90^(@)=156^(@)`

Discussion

5.	The radius of curvature of curved surface of a thin plano-convex lens is `10 cm` and the refractive index is `1.5`. If the plano surface is silvered, then the focal length will be.A. 15 cmB. 20 cmC. 5 cmD. 10 cm
Answer» Correct Answer - D

Discussion

6.	Calculate the resolving power of a microscope with cone angle of light falling on the objective equal to `60^(@)`. Take `lambda = 600 nm mu` for air `= 1`.
Answer» Here, `2 theta = 60^(@), theta = 30^(@)`, `lambda = 600 nm = 6 xx 10^(-7) m , mu = 1` `R.P. = (1)/(d) = (2 mu sin theta)/(lambda)` `R.P. = (2 xx 1 xx sin 30^(@))/(600 xx 10^(-9)) = (10)/(6) xx 10^(6) = 1.67 xx 10^(6)`

Discussion

7.	The diameter of the pupil of human eye is about `2 mm`. Human eye is most sensitive to the wavelength `555 nm`. Find the limit of resolution of human eye.
Answer» Here, `D = 2 mm = 2 xx 10^(-3)m` `lambda = 555 nm = 555 xx 10^(-9)m` Limit of resolution, `dtheta = (1.22 lambda)/(D) = (1.22 xx 555 xx 10^(-9))/(2 xx 10^(-3))` `= 3.39 xx 10^(-4)rad` `dtheta = 3.39 xx 10^(-4) xx ((180)/(pi))^(@)` `= 0.0194^(@) = 0.0194 xx 60 min = 1.2 min`

Discussion

8.	Assume that light of wavelength `6000 Å` is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch ? (NCERT Solved example)
Answer» Diameter of onjective of telescope, `D = 100 "inch" = 100 xx 2.54 cm = 254 cm` `lambda = 6000 Å = 6000 xx 10^(-8)cm = 6 xx 10^(-5) cm` Limit of resolution of telescope, `d theta = (1.22 lambda)/(D) = (1.22 xx 6 xx 10^(-5))/(254)` `= 2.9 xx 10^(-7)` radian

Discussion

9.	A central fringe of interference pattern produced by light of wavelength `6000 Å` is shifted to the position of 5th bright fringe by introducing thin film of `mu = 1.5`. Calculate thickness of the film.
Answer» Correct Answer - 6 micron Here, `lambda = 6000Å = 6 xx 10^(-7) m` `n = 5, mu = 1.5, t =?` The basic relation used is that path difference introduced by film = Path diff. for the shift. `(mu - 1) t = n lambda` `t = (n lambda)/((mu - 1)) = 6 xx 10^(-6)m` `= 6 micron`

Discussion

10.	Laser light of wavelength `630 nm` incident on a pair of slits produces an interference pattern where bright fringes are separated by `8.1 mm`. Another light produces the interference pattern, where the bright fringes are separated by `7.2 mm`. Calculate the wavelength of second light.
Answer» Here, `lambda_(1) = 630 nm, beta_(1) = 8.1 mm` As `beta = ( lambda D)/(d)`, therefore, `(lambda_(2))/(lambda_(1)) = (beta_(2))/(beta_(1)) = (7.2)/(8.1) = (8)/(9)` `lambda_(2) = (8)/(9) lambda_(1) = (8)/(9) xx 630 nm = 560 nm`

Discussion

11.	Two lenses of power +12D and -2D are combined together. What is their equivalent focal length?(a) 10 cm (b) 12.5 cm (c) 16.6 cm (d) 8.33 cm
Answer» (a) 10 cm P = P₁ + P₂ = + 12 – 2 = 10 D F = \(\frac{1}{P}\) = \(\frac{1}{10}\) m = 10 cm

11.

Two lenses of power +12D and -2D are combined together. What is their equivalent focal length?(a) 10 cm (b) 12.5 cm (c) 16.6 cm (d) 8.33 cm

Answer»

(a) 10 cm

P = P₁ + P₂ = + 12 – 2 = 10 D

F = \(\frac{1}{P}\) = \(\frac{1}{10}\) m = 10 cm

Discussion

12.	Why is oil immersed objective preferred in a microscope?
Answer» It is best to use an oil-immersed objective at high magnification as the oil compensates for short focal lengths associated with larger magnifications.

Discussion

13.	A ray `PQ` incident on face `AB` of a prism `ABC`, as shown in Fig., emerges from the face `AC` such that `AQ = AR`. Draw the ray diagram showing the pasage of the ray through the prism. If the angle of prism is `60^@` and refractive index of the material of the prism is `sqrt(3) ,` determine the values of angle of incidence and angle of deviation. .
Answer» As the refracted ray passes such that `AQ = AR`, it must go parallel to the base of the prism suffring minimum deviation. `r = A//2 = (60)/(2) = 30^(@)` `mu = (sin i)/(sin r)/(sin i)/(sin 30^(@))` `:. sin I = mu sin 30^(@) = sqrt(3) xx (1)/(2)` `i = 60^(@)` As `i + e =A + delta_(m)` `2i = 60^(@) + delta_(m), delta_(m) = 2 i - 60^(@) = 120^(@) - 60^(@)` `delta_(m) = 60^(@)`

Discussion

14.	Why is yellow light preferred to during fog?
Answer» Yellow light has longer wavelength than green, blue or violet components of white lights. As scattered intensity, I ∝ \(\frac{1}{λ^4}\) . so yellow colour is least scattered and produces sufficient illumination.

14.

Why is yellow light preferred to during fog?

Answer»

Yellow light has longer wavelength than green, blue or violet components of white lights. As scattered intensity, I ∝ \(\frac{1}{λ^4}\) . so yellow colour is least scattered and produces sufficient illumination.

Discussion

15.	One can not see through fog, because (a) fog absorbs the light (b) light suffers total reflection at droplets (c) refractive index of the fog is infinity (d) light is scattered by the droplets
Answer» (d) light is scattered by the droplets

Discussion

16.	One cannot cannot see through fog, becauseA. fog absorbs the lightB. light suffers total reflection at dropletsC. refractive index of fog is infinityD. light is scattered by droplets
Answer» Correct Answer - D

Discussion

17.	A ray of light is inclined to one face of a prism at an angle of `60^@`. If angle of prism is `60^@` and the ray deviated through an angle of `42^@` find the angle which the emergent ray makes with second face of the prism.
Answer» Correct Answer - `18^(@)` Here, `i_(1) = 90^(@) - 60^(@) = 30^(@), A = 60^(@)`, `delta = 42^(@) , (90 - i_(2)) = ?` Use `i_(1) + i_(2) = A + delta` Remember that `i_(1), i_(2)` are angles with normal to the faces of the prism.

Discussion

18.	For a single slit of width `a`, the first minimum of diffraction pattern of a monochromatic light of wavelength `lambda` occurs at an angle of `lambda//a`. At the same angle of `lambda//a`, we get a maximum for two narrow slits separated by a distance a. Explain.
Answer» In the interference pattern due to narrow slits separated by distance `a`, path differecne `= a sin theta`. For constructive interference, `a sin theta = 1 lambda, sin theta ~= theta = lambda//a`, we get a maximum. In the diffraction pattern due to a single slit, when path difference from secondary waves from the ends of wavefront is `lambda`, then for every point in one half of the lower half for which path diff. between secondary waves is `lambda//2`. Therefore, interference is destructive and we obtain first minimum at the same angle.

Discussion

19.	How wide is the base of a trihedral prism which has the same resolving power as a diffraaction grating with `10000` lines in the second order of the spectrum if `\|dn//d lambda\| = 0.10mu m^(-1)`?
Answer» `\|(dn)/(d lambda)\| = kN = 2 xx 10,000` `b xx 0.10 mu m^(-1) = 2 xx 10^(4)` `b= 2xx 10^(5)mum = 0.2m = 20cm`.

Discussion

20.	There is a telescope whose objective has a diameter `D = 5.0 cm`. Find the resolving power of the objective and the minimum separation between two points at a distance `l = 3.0km` from the telescope, which it can resolve (assume `lambda = 0.55 mu m)`.
Answer» Resolving power of the objective `=(D)/(1.22 lambda) = (5 xx 10^(-2))/(1.22 xx 0.55 xx 10^(-6)) = 7.45 xx 10^(4)` Let `(Deltay)_(min)` be the minimum distance between two points at a distance of `3.0km` which the telescope can resolve. Then `((Deltay)_(min))/(3 xx 10^(3)) = (1.22 lambda)/(D) = (1)/(7.45 xx 10^(4))` or `(Deltay)_(min) = (3 xx 10^(3))/(7.45 xx 10^(4)) = 0.04026m = 4.03cm`.

Discussion

21.	Estimate the distance for which ray optics is good approximately for an aperture of 2mm and wavelength 500 nm.
Answer» Ray optics is good approximation upto Fresnel distance, `Z_(F) = (a^(2))/(lambda) = (4 xx 10^(-3))^(2)/(400 xx 10^(-9)) = (16 xx 10^(-6))/(4 xx 10^(-7)) = 40m`

Discussion

22.	Assertion : The resolving power of a telescope is more if the diameter of the objective lens is more. Reason : Objective lens of large diameter collectd more light.A. If both, Assertion and Reason are true and the Reason is the correct explaination of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explaination of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false.
Answer» Correct Answer - A Both, the Assertion and Reason are true and latter explains the former correctly. As `R.P. = (D)/(1.22 lambda)`, where `D` is the diameter of the objective lens. Moreover, large objective lens collects more light and hence image becomes brighter.

Discussion

23.	What is the relation between magnifying power and resolving power of a telescope ?
Answer» Magnifying power `= ("Resolving power of eye")/("Resolving power of telescope")`.

Discussion

24.	The focal length of a thin lens in vacuum is `f`. If the material of the lens has `mu = 3//2`, its focal length when immersed in water of refractive index `4//3` will be.A. `f`B. `4 f//3`C. `2 f`D. `4 f`
Answer» Correct Answer - D `(1)/(f_(a)) = ((mu_(g))/(mu_(a)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` …(i) `(1)/(f_(w)) = ((mu_(g))/(mu_(w)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` …(ii) Dividing (i) by (ii), we get `(f_(w))/(f_(a)) = ((mu_(g)//mu_(a) - 1))/((mu_(g)//mu_(w) - 1)) = ((3//2 - 1))/(((3//2)/(4//3) -1)) = (1//2)/(1//8) = 4` `:. f_(w) = 4 f_(a) = 4 f`

Discussion

25.	A convex mirror of focal length f (in air) is immersed in water `(mu = 4/3)`. The focal length of the mirror in water will be -A. `f`B. `(4/3)f`C. `(3/4)f`D. `(7/3)f`
Answer» Correct Answer - A

Discussion

26.	Image formed by a concave mirror radius of curvature 40 cm is half the size of the object. Then distance of object and its image from the mirror will be -A. `30 cm` and `60 cm`B. `60 cm` and `120 cm`C. `60 cm` and `30 cm`D. `120 cm` and `60 cm`
Answer» Correct Answer - C

Discussion

27.	Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is(A) 4 × 10-11 s(B) 2 × 10-11 ns(C) 16 × 10-11 s(D) 8 × 10-10 s
Answer» Correct option is: (A) 4 × 10^-11 s

Discussion

28.	Calculate the time taken by light to travel a distance of `10 km` in water of refractive index `4//3`.
Answer» Here, `t = ? S = 10 km = 10^(4)m` `mu = 4//3` `t = (s)/(v) = (s)/(c//mu) = ((4)/(3) xx 10^(4))/(3 xx 10^(8))s` `= 4.44 xx 10^(-5)s`

Discussion

29.	The refractive index of diamond is `2.47` and that of window glass is `1.51`. Find the ratio of speed of light in glass and diamond.
Answer» Here, `mu_(d) = 2.47 and mu_(g) = 1.51` As `mu = c//v :. (v_(g))/(v_(d)) = (mu_(d))/(mu_(g)) = (2.47)/(1.51) = 1.63`

Discussion

30.	Which one has a greater critical angle diamond or glass ?
Answer» `.^(a)mu_(D) = 2.46` and `.^(a)mu_(g)`. As `mu = (1)/(sin C)`, therefore, `C_(D) lt C_(g)`

Discussion

31.	The magnifying power of an astronomical telescope in the normal adjustment position is `100`. The distance between the objective and eye piece is `101 cm` . Calculate the focal lengths of objective and eye piece.
Answer» Correct Answer - `100 cm` and `1 cm` `m = - 100, f_(0) + f_(e) = 101 cm, f_(0) = ?, f_(e) = ?` `m = -(f_(0))/(f_(e)) = - 100 :. F_(0) = 100 f_(e)` Now `f_(0) + f_(e) = 101` `100 f_(e) + f_(e) = 101`, `f_(e) = 1 cm` `f_(0) = 100 f_(e) = 100 cm`

Discussion

32.	Two boys one 52 inches tall and the other 55 inches tall, are standing at distances 4.0 m and 5.0 m respectivley from an eye. Which boy will taller?
Answer» Correct Answer - First boy Angle subtended by the first boy, on the eye `alpha_(1) = (52 I nch)/(4.0 m) = 13 "inch"//m` Angle subtended by the second boy on the eye is `alpha_(2) = (55 "inch")/(5.0 m) = 11 "inch"//m` As `alpha_(1) gt alpha_(2)`, therefore the first boy will look taller to the eye.

Discussion

33.	A telescope consists of two lenses of focal lengths `0.3 m and 3 cm` respectively. It is fucussed on moon which subtends an angle of `0.5^@` at the obejctive. Calculate the angle subtended at the eye by the final image in normal adjustment of the telescope.
Answer» Correct Answer - `5^(@)` Here, `f_(0) = 0.3 m` and `f_(e) = 3 cm, alpha = -0.5^(@), beta = ?` Use `(beta)/(alpha) = (f_(0))/(f_(e))`

Discussion

34.	What is the angle between the reflected and refracted rays at polarizing angle ?
Answer» Correct Answer - `90^(@)`.

Discussion

35.	What is the value of refractive index of a mediumof polarizing angle `60^(@)` ?
Answer» `mu = tan i_(p) = tan 60^(@) = sqrt(3)`.

Discussion

36.	When sun light is incident at an angle of `53^(@)` on the surface of water, the reflected light is plane polarized. Calculate the angle of refraction and refractive index of water.
Answer» Correct Answer - `37^(@), 1.327`

Discussion

37.	The refractive index of a medium is `sqrt(3)`. What is the angle of refraction, if polarizing angle of the medium.
Answer» Here, `mu = sqrt(3) , r = ?` As `tan i_(P) = mu = sqrt(3)` `:. i_(P) = tan^(-1) (sqrt(3)) = 60^(@)` As `r = 90^(@) - i_(P) :. R = 90^(@) - 60^(@) = 30^(@)`

Discussion

38.	Critical angle for a certain wavelength of light in glass is `40^(@)`. Calculate the polarizing angle and the angle of refraction in glass corresponding to it.
Answer» Correct Answer - `57.3^(@), 32.7^(@)`

Discussion

39.	A small air bubble in glass shpere of radius `2 cm` appears to be `1 cm` from the surface when looked at, along a diameter. If the refractive index of glass is `1.5`, find the true position of the air bubble.
Answer» Correct Answer - `1.2 cm` Here, `mu_(1), mu_(2) = 1.5, R = - 2cm` Incident ray `OA` in glass is refracted in air, along `AB`, and appears to some from `I`. `u = PO = ? v = PI = - 1 cm` As refraction occurs from denser to rarer medium, `:. -(mu_(2))/(u) + (mu_(1))/(v) = (mu_(1) - mu_(2))/(R )` `-(1.5)/(u) + (1)/(-1) = (1 - 1.5)/(-2) = (1)/(4)` `(1.5)/(u) = -1 - (1)/(4) = -(5)/(4)` `u = (-4 xx 1.5)/(5) = - 1.2 cm` The air bubble `O` lies at `1.2 cm` from the refracting surface within the sphere.

Discussion

40.	An air bubble is inside water. The refractive index of water is 4/3. At what distance from the air bubble should a point object be placed so as to form a real image at the same distance from the bubble? (a) 2R (b) 3R (c) 4R (d) The air bubble cannot form a real image.
Answer» Correct Answer is: (d) The air bubble cannot form a real image.

Discussion

41.	What type of lens is formed by a bubble inside water?
Answer» Air bubble has spherical surface and is surrounded by medium (water) of higher refractive index. When light passes from water to air it gets diverged. So air bubble behaves as a concave lens.

Discussion

42.	An air bubble is inside water. The refractive index of water is `4//3`. At the what distance from the air bubble should a point object be placed so as to form a real image at the same distance from the bubble?A. `2R`B. `3R`C. `4R`D. The air bubble cannot from a real image.
Answer» Correct Answer - D

Discussion

43.	What type of lens is an air bubble inside water ?
Answer» it behaves as a concave lens because water is denser than air.

Discussion

44.	What type of lens is a tumbler filled with water ?
Answer» A trumbler filled with water behave as a double convex lens.

Discussion

45.	There are different fish, monkeys and water of the habitable planet of the star Proxima b. A fish swimming underwater feels that there is a monkey at 2.5 m on the top of a tree. The same monkey feels that the fish is 1.6 m below the water surface. Interestingly, height of the tree and the depth at which the fish is swimming are exactly same. Refractive index of that water must be (A) 6/5 (B) 5/4 (C) 4/3 (D) 7/5
Answer» Correct answer is (B) 5/4

Discussion

46.	When a fish looks up the surface of a perfectly smooth lake, the surface appears dark except inside a circular area directly above it. Calculate the angle that this illuminated region subtends. Given `mu` of water `= 1.333`.
Answer» Correct Answer - `97.2^(@)` Here, `mu = 1.333, 2 theta = ?` `sin C = (1)/(mu) = (1)/(1.333) = 0.75` `C = sin^(-1)(0.75) = 48.6^(@)`. Angle which illuminated region subtends `= 2 theta = 2 C = 2 xx 48.6^(@) = 97.2^(@)`

Discussion

47.	A plane monochromatic light wave with intensity `I_(0)` falls normally on an opaque disc closing the first Fresnel zone for the observation point `P`. What did the intensity of light `I` at the point `P` become equal to after (a) half of the disc (along the diameter) was removed, (b) half of the external half of the first Fresnal zone removed (along the diameter)?
Answer» Suppose the disc does not abstruct light at all. Then `A_(disc) + A_("remainder") = (1)/(2)A_(disc)` (because the disc covers the first Fresnel zone only). So `A_("remainder") =- (1)/(2)A_(disc)` Hence the amplitude when half of the disc is removed along a diameter `= (1)/(2)A_(disc) + A_("remainder") = (1)/(2)A_(disc) - (1)/(2)A_(disc) ~~0` Hence `I = 0`. (b) In this case `A = (1)/(2)A_("external") +A_("remainder")` `= (1)/(2)A_("external") - (1)/(2)A_(disc)` We write `A_(disc) = A_(in) + iA_(ou)` where `A_(in) (A_(out))` stands for `A_("internal") (A_("external"))`. The factor `i` takes account of the `(pi)/(2)` pahse difference between two halves of the first Fresnel zone. Thus `A =-(1)/(2)A_("in")` and `I = (1)/(4)A_("in")^(2)` On the other hand `I_(0) = (1)/(4) (A_("in")^(2) + A_(out)^(2)) = (1)/(2) A_("in")^(2)` so `I = (1)/(2)I_(0)`.

Discussion

48.	In a double slit experiment, two coherent sources have slightly different intensities `I` and `(I + delta I)`, such that `delta I lt lt I`. Show that resultant intensity at maxima is near `4 I`, while that at minima is nearly `(delta I)^(2)//4 I`.
Answer» From `I = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos phi` `I_(max) = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos 0^(@) = I + (I + delta I) + 2 sqrt(I(I + delta I))` As `deltaI lt lt I`, therefore, `I_(max) I + I 2 I = 4I` Again from `I = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos phi` `I_(min) = I + (I + deltaI) + 2sqrt(I(I + deltaI)) cos 180^(@) = 2I + delta I - 2 I(1 + (delta I)/(I))^(1//2)` `= 2 I + delta I - 2 I [1 + (1)/(2) (delta I)/(I) + ((1)/(2)((1)/(2) - 1))/(2!)((delta I)/(I))^(2)] = 2 I + delta I - 2 I - delta I + (1)/(4)I((delta I)/(I))^(2) = (delta I)^(2)/(4I)`

Discussion

49.	A plane monochromatic light wave with intensity `I_(0)` falls normally on the surfaces of the opaque screens shown in Fig. Find the intensity of light `I` at a point `P` (a) located behind the corner ponts of screens `1-3` and behind the edge of half-plane `4`, (b) for which the rounded-off edge of screens `5-8` coincides with the boundary of the first formula describing the result obtained for screens `1-4`, the same, for screens `5-8`.
Answer» When the screen is fully transparent, the amplitude of vibrations is `(1)/(2)A_(1)` (with intensity `I_(0) = (1)/(4)A_(1)^(2))`. (a) `(1)` In this case `A = (3)/(4) ((1)/(2)A_(1))` so squaring `I = (9)/(16)I_(0)` `(2)` In this case `(1)/(2)` of the plane s blacket out so `A = (1)/(2) ((1)/(2)A_(1))` and `I = (1)/(4)I_(0)` `(3)` In this case `A = (1)/(4) (A_(1)//2)` and `I = (1)/(16)I_(0)`. `(4)` In this case `A = (1)/(2) ((1)/(2)A_(1))` again and `I = (1)/(4)I_(0)` so `I_(4) = (I)/(2)` In general we get `I(varphi) = I_(0) ((2varphi)/(2pi)))^(2)` where `varphi` is the total angle blocket out by the screen. `(b) (5)` Here `A= (3)/(4) ((1)/(2)A_(1)) + (1)/(4)A_(1)` `A_(1)` being the contribution of the first Fresnel zone. Thus `A = (5)/(8)A_(1)` and `I = (25)/(16)I_(0)` `(6) A = (1)/(2) ((1)/(2)A_(1)) + (1)/(2)A_(1) = (3)/(4)A_(1)` and `I = (9)/(4)I_(0)` `(7) A = (1)/(4) ((1)/(2)A_(1)) + (3)/(4)A_(1) = (7)/(8)A_(1)` and `I = (49)/(16)I_(0)` `(8) A = (1)/(2) ((1)/(2)A_(1)) + (1)/(2)A_(1) = (3)/(4)A_(1)` and `I = (9)/(4)I_(0) (I_(8) - I_(6))` In `5` to `8` the first term in the expression for the amplitude is the contribution of the plane part and the second term gives the expression for the Fresnel zone part. in general in `(5)` to `(8) I = I_(0) (1+ ((varphi)/(2pi))^(2))` when `varphi` is the angle convered by the screen.

Discussion

50.	A plane monochromatic light wave of intensity `I_(0)` falls normally on an opaque screen with a long slit having a semicircular cut on one side (Fig.) The edge of the cut coincides with the boundary line of the first Fresnel zone for the observation point `P`. Thw width of the slit measure `0.90` of the radius of the cut. using Fig. find the intensity of light at the point `P`.
Answer» we apply the formula of problem `5.103` and calculate `underset("aperture")int (a_(0))/(r ) e^(-ikr) dS = underset("semicircle")int + underset("slit")int` The contribution of the full `1^(st)` Fresnel zone has been evaluated in `5.103`. The contribution the semi-circule is one half of it and is `-(2pi)/(k)i a_(0)e^(-ikb) =- ia_(0) lambda e^(-ikb)` The contribution of the slit is `(a_(0))/(b) underset(0)overset(0.90sqrt(b lambda))int e^(-ikb) e^(-ik(x^(2))/(2b)) dx underset(-oo)overset(oo)int e^(-iky^(2)//2b) dy` Now `underset(-oo)overset(oo)int e^(-iky^(2)//2b) dy = underset(-oo)overset(oo)int e^(-i(piy^(2))/(b lambda)) dy` `sqrt((b lambda)/(2)) underset(-oo)overset(oo)int e^(-ipiy^(2)//2) du = sqrt(b lambda) e^(-ipi//4)` Thus the contribution of the slit is `(a_(0))/(b) sqrt(b lambda) e^(-ik b-i pi//4) underset(-oo)overset(0.9 xx sqrt(2))int e^(-i piu^(2)//2) du sqrt((b lambda)/(2))` `= a_(0)lambda e^(-ikb-ipi//4) (1)/(sqrt(2)) underset(0)overset(1.27)int e^(-ipiu^(2)//2) du` Thus the intensity at the observation point `P` on the screen is `a_(0)^(2)lambda^(2) \|-i+(1 - i)/(2)(C(1.27) - iS(1.27))\|^(2) = a_(0)^(2)lambda^(2) \|-i + ((1-i)(0.67 - 0.65i))/(2)\|^(2)` (on using `(C(1.27) = 0.67` and `S(1.27) = 0.65)` `= a_(0)^(2) lambda^(2)\|-i + 0.01 - 0.66i\|^(2)` `a_(0)^(2)lambda^(2)\|0.01 - 1.66i\|^(2)` `= 2.76 a_(0)^(2)lambda^(2)` Now `a_(0)^(2)lambda^(2)` is the intensity due to half of `1^(st)` Fresnel zone and is therefore equal to `I_(0)`. (It can also be obtained by doing the `x`-intergal over `-oo` to `+oo`). Thus `I = 2.76 I_(0)`.

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