Wave Optics Class 12th Physics Part Ii CBSE Solution

Class 12th Physics Part Ii CBSE Solution
Exercises
  1. Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are…
  2. What is the shape of the wave-front in each of the following cases: (a) Light diverging…
  3. The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light…
  4. Is the speed of light in glass independent of the colour of light? If not, which of the…
  5. In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is…
  6. In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity…
  7. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain…
  8. In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen…
  9. What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)…
  10. Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength…
  11. Estimate the distance for which ray optics is good approximation for an aperture of 4 mm…
Additional Exercises
  1. The 6563 Å Hα line emitted by hydrogen in a star is found to be red shifted by 15 Å.…
  2. Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be…
  3. You have learnt in the text how Huygens’ Principle leads to the laws of reflection and…
  4. Let us list some of the factors, which could possibly influence the speed of wave…
  5. For sound waves, the Doppler formula for frequency shift differs slightly between the two…
  6. In double - slit experiment using light of wavelength 600 nm, the angular width of a…
  7. In a single slit diffraction experiment, the width of the slit is made double the original…
  8. In what way is diffraction from each slit related to the interference pattern in a double…
  9. When a tiny circular obstacle is placed in the path of light from a distant source, a…
  10. Two students are separated by a 7 m partition wall in a room 10 m high. If both light and…
  11. Ray optics is based on the assumption that light travels in a straight line. Diffraction…
  12. Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a…
  13. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting…
  14. When a low flying aircraft passes overhead; we sometimes notice a slight shaking of the…
  15. As you have learnt in the text, the principle of linear superposition of wave displacement…
  16. In deriving the single slit diffraction pattern, it was stated that the intensity is zero…

Exercises
Question 1.

Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.


Answer:

Given:

Wavelength of incident light (λ) = 589 nm


Refractive Index of water = 1.33


λ = Wavelength of light


c = Speed of light in vacuum/ air = 3 × 108 m/s


f = Frequency of light



Figure showing reflection and refraction of a light ray


(a) In case of Reflection:


Explanation: Wavelength and frequency of light do not change in case of reflection as the speed of light remains constant because the reflected light is in the same medium as that of incident light.


Wavelength of reflected light (λ1) = 589 nm


Frequency of reflected light (f1) =  = 


= 5.093 × 1014 Hz


Speed of reflected light (c1) = f1λ1 = 3 × 108 m/s


Wavelength, frequency and speed of reflected light are 589nm, 5.093 × 1014Hz and 3 × 108 m/s respectively.


(b) In case of Refraction:


Explanation:In refraction velocity of light changes as it enters another medium. As a result its wavelength changes but frequency remains constant.


v = Velocity of light in a particular medium


μ = Refractive index of medium = 


So, μ = 


⇒ 


⇒ v2 = 2.25 × 108 m/s.


We know that f2 = f1 = 5.093 × 1014 Hz



⇒ 


Or λ = 442.8nm


Wavelength, frequency and speed of refracted light are 442.8nm, 5.093 × 1014 Hz and 2.25 × 108 m/s respectively.



Question 2.

What is the shape of the wave-front in each of the following cases:

(a) Light diverging from a point source.

(b) Light emerging out of a convex lens when a point source is placed at its focus.

(c) The portion of the wave-front of light from a distant star intercepted by the Earth


Answer:

Wave-front :


The locus of all the particles of the medium vibrating in the same phase at a given instant is called wave-front.


(a) In case of light diverging from a point source, the shape of wave-front is a circle



Figure of wave-front formed by a point source


(b) In case of light emerging out of a convex lens when a point source is placed at its focus, the wave-front is a parallel grid.



Fig: Light rays passing through a convex lens



Figure of wave-front formed by a convex lens when point source is at its focus


(c) When light source is a distance star intercepted by earth, the wave-front formed is a plane 



Question 3.

The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 × 108 m s–1)


Answer:

Given:


Refractive index of glass (μ) = 1.5


v = Velocity of light in a medium


Now, the Refractive index of medium is given as,



According to the formula,



v = 2 × 108 m/s.


Speed of light in glass is 2 × 108 m/s.



Question 4.

Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?


Answer:

Speed of light in glass is dependent on the colour of light as different colours have different wavelengths, thus different velocities.


We know that refractive index of a material in terms of the wavelength of light is


μ = a + (b/λ2)


As the wavelength of violet light is smaller than that of red light,


μv > μr.


We also know that 


Or 


⇒ vv < vr


Hence velocity of violet light in glass is less than velocity of red light.



Question 5.

In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.


Answer:


Figure showing bright fringes in Young's double slit experiment


Given:


Distance between the slits (d) = 0.28 mm


Distance between the slits and the screen (D) = 1.4 m


Distance between the central bright fringe and fourth bright fringe (yn) = 1.2 cm


where n is the order of the fringe, here, n = 4


λ = Wavelength of light used


From the formula,




λ = 600 nm
Hence ,The wavelength of light used is 600 nm


Question 6.

In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?


Answer:

Given:


Wavelength of monochromatic light = λ


Intensity of light at a point on the screen where path difference is λ = K units.


Intensity of light at any point = I


Intensity of light from slit 1 = I1


Intensity of light from slit 2 = I2


Initially let us consider I1 = I2 = I0


When path difference is λ, the phase difference is 2


Using the formula to find out intensity of resultant wave,


I = I1 + I2 + 2√(I1I2) cos(θ)


k = I0 + I0 + 2 cos π = 4I0 (∵ Cos(2π) = 1)


k = 4I0


Or, I0 = k/4


When path difference is λ/3, the phase difference is 2/3


I = I1 + I2 + 2√(I1I2) cos(θ)


I = I0 + I0 + 2√(I1I2) cos(2π/3)I0 + I0 + 2I0 × (-1/2) = I0 = k/4


Intensity of light when path difference is λ/3 is k/4.



Question 7.

A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?


Answer:

Distance of nth fringe from central maxima, Xn = nλD/d


Where, λ is wavelength of light


D is distance from slits to screen


d is slit width


(a) We need to find distance of 3rd from central maxima for wavelength 650 nm


n = 3


x3 = 3 × (650 nm) D/d


(b) The distance of nth bright fringe from central maxima for two wavelengths, say λ1 and λ2, are


Xn = nλ1D/d


Yn = nλ2D/d


When bright fringe due to two wavelength coincide, their distance from central maxima is same


i.e., Xn = Yn


n1λ1D/d = n2λ2D/d


n1λ1 = n2λ2


n1/n2 = λ2 /λ1


= 520 / 650 = 8/10


Therefore, the bright fringe due to two wavelength coincide when n1 is integer multiple of 8 and n2 is integer multiple of 10.


Thus least distance for which they coincide,


S = n1 λ1D/d


= 8 × (650 nm) D/d


[The question is not complete. This question cannot be answered completely without the values of d and D.]



Question 8.

In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.


Answer:

nth order Angular width of fringe in air,


sin(θn) = nλ/d … (1)


Where sin(θ) is angular fringe width


n is order of fringe


λ is wavelength of light used


d is slit width


When entire apparatus is taken into water, d and D remain same


∴ Angular fringe width for nth order in water,


sin(θnw) = nλw/d … (2)


where λw is wavelength of the light in water


Taking ratio of equations (1) & (2)


sin(θn)/sin(θnw) = λ/ λw


Refractive index of water is 4/3


i.e., 4/3 = c/v


Where c is velocity of light in air (vacuum) and v is velocity in water


c/v = ν λ /ν λw


Since frequency does not vary with medium, ν = νw


Thus 4/3 = λ/λw


Thus we have,


sin(θ)/sin(θw) = 4/3


sin(θw) = sin(θ) × 3/4


= 0.2 × 3/4 = 0.15°



Question 9.

What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)


Answer:

Brewster’s angle is angle of incidence of light on a refracting surface for which refracted and reflected light are linearly polarised. They are also perpendicular to each other.



According to Brewster’s law when angle of incidence is Brewster angle, say θB,


tan(θB) = μ


where μ is refractive index of the refracting medium


i.e., tan(θB) = 1.5


θ = tan-1(1.5)


= 56.3°



Question 10.

Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?


Answer:

During reflection at a surface, wavelength and frequency of light wave remain same, only phase of the wave changes by 180°.


Thus, wavelength of reflected light is 5000 Å itself.


We know c = ν λ


where c, ν, λ are velocity, frequency, wavelength respectively.


c = ν × 5000 × 10-10


∴ ν = 3 × 108/5000 × 10-10


= 6 × 1014 Hz


Wavelength of reflected light is 5000 Å


and frequency is 6 × 1014 Hz


Angle between incident and refracted ray,


θ = I + R



Where ‘I’ is incident angle and ‘R’ is reflected angle.


We need to find angle of incidence for which angel between incident and reflected ray is 90°.


i.e., θ = 90°


I + R = 90°


From laws of reflection,


Angle of incidence = Angle of reflection


i.e., I = R


∴ I + I = 2I = 90°


I = 45°



Question 11.

Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.


Answer:

An aperture illuminated by beam of light sends diffracted light. The distance travelled by diffracted beam at which, width of spread of central maxima (zλ/a) equals the size of the aperture (a) is called the Fresnel distance(ZF). Beyond this distance, ray optics is good approximation for an optical measurement.


For a light wave of given wavelength, say λ, and an aperture of length ‘a’ length, Fresnel distance,


ZF = a2


a = size of aperture


λ = wavelength of light


Given,


a = 4 mm = 4 × 10-3 m


λ = 400 × 10-9 m


Thus, ZF = (4 × 10-3)2 / 400 × 10-9


= 40 m


For distance greater than 40 m, ray optics is valid.




Additional Exercises
Question 1.

The 6563 Å Hα line emitted by hydrogen in a star is found to be red shifted by 15 Å. Estimate the speed with which the star is receding from the Earth.


Answer:

Redshift is a Doppler’s effect phenomena seen in the case of electromagnetic waves, It happens whenever the light source moves away from the observer or vice versa and the light or other electromagnetic radiation from an object seems to be increased in wavelength, or shifted to the red end of the spectrum. The wavelength of the light is observed increased because of the change in the frequency of the light observed, since the departing source will give an impression of decreased frequency or increased wavelength, the light observed will shift towards red spectrum or the larger wavelength.


The relation of red shift with velocity is



Where,


λ’ is the apparent increased wavelength,


λ is the original wavelength


δλ is the difference between apparent and original wavelength.


Given:


Wavelength of Hα line emitted by hydrogen


λ = 6563 Å = 6563 × 10 -10 m.


Red shift of the star, δλ = (λ’ - λ) = 15 Å = 15 × 10- 10 m


Speed of Light, c = 3 × 108 m/s


We suppose the velocity of the star receding away from the Earth as v.


Since, the ratio of apparent change in the wavelength is equal to the ratio of velocity of receding body and the standard speed of light.




⇒ v = 6.87 × 105 m/s


The estimatedspeed with which the star is receding from the Earth is 6.87 × 105 m/s



Question 2.

Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?


Answer:

Newton’s Corpuscular theory of light theorised the light to consist of small corpuscles and states that, when light corpuscles strike the interface of two media from a rarer (here air) to denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, component of the velocity normal to the point of interaction with the surface increases while the component parallel along the surface remains unchanged. Thus component of velocity of light in one direction will be equal.


Suppose we take


v as the velocity of the light in rarer medium (air)


v’ as velocity in the denser medium (water)


The components parallel to the surfaces can be written as,


v sini and v’sinr


where,


i is the angle of incidence in the rarer medium,


r is the angle of refraction in the denser medium


But, v sin(i) = v’ sin(r)


⇒ 


(where μ is the refractive index). And, we have refractive index ‘μ’ always greater than unity.


Thus, light should travel faster in the denser medium like water than in air.


Note: The experimental results prove that no object can travel faster than the speed of light i.e. c.


Thus,


No, the predictions of Newton’s Corpuscular theory were not confirmed by any experimental determination of the speed of light in water.


The Huygens wave theory predicts that v’ is greater than v which is consistent with the experimental results.



Question 3.

You have learnt in the text how Huygens’ Principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.


Answer:

Huygens principle is used to explain phenomena such as refraction, reflection and diffraction by describing the light to travel forward in the form of wave fronts. These primary wave fronts act as source for the next secondary one in the propagation of light.


We start by supposing that an object O placed in front of a plane mirror MP at a distance equal to the radius of the circle (as shown in the figure below)



A circle is drawn from the centre taken as object O such that the plane mirror is just tangent to the circle at the point P. The Huygens’s Principle thus points that AB is the first wave front of the incident light from the object O.


If we suppose the absence of the mirror, then by Huygens principle a similar wave front can be drawn parallel to AB behind the P at a distance of the same radius of the circle (as given in the figure below).


A’B’ can be considered as a virtual reflected ray for the plane mirror. Therefore, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object at an equal distance. Further another secondary wave front can be drawn by taking the circle A’B’ and placing a mirror tangent to the circle and parallel to the previous one. The point ‘P’ will act as an object for the next virtual image. The secondary wave fronts thus are at a distance of the radius of the circle from the tangent of a virtual plane mirror.



Question 4.

Let us list some of the factors, which could possibly influence the speed of wave propagation:

(i) Nature of the source.

(ii) Direction of propagation.

(iii) Motion of the source and/or observer.

(iv) Wavelength.

(v) Intensity of the wave.

On which of these factors, if any, does

(a) The speed of light in vacuum,

(b) The speed of light in a medium (say, glass or water), depend?


Answer:

(a) The speed of light in vacuum i.e. 3 × 108 m/s is universal constant and thus, it’s value stays the same in vacuum as there are no interacting particles, It is not effected by (i) the nature of the source, (ii) Direction of propagation (iii) Motion of the source or observer (iv) Wavelength and (v) Intensity of the wave.


Thus, the given factors will have no effect on the speed of light in vacuum.


(b) Since, the factors such as (i) the nature of the source (ii) the direction of propagation (iv) wavelength change the intrinsic property of the light but other factors given are external factors and does not affect the propagation property of light. Thus, The speed of light in a medium like glass or water is influenced only by (i) (ii), (iv).



Question 5.

For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?


Answer:

(i) No, the properties of propagation of light and sound are very different from each other. Sound waves can propagate only in the presence of a medium. The velocities do not just simply add up when they approach the speed of light.


(ii) The two situations whether it is source at rest or observer at rest are not identical because the sound waves are mechanical and thus depends on the properties of the medium through which it is travelling.


In case of the observer moving the apparent change in the frequency can be given as



In case of source moving the apparent change frequency can be given as,



Where,


f is the frequency of waves from source


f’ is the apparent frequency observed.


ν’ is the speed of the observer and ν is the speed of the wave from the source


The negative sign is for the observer or source moving away from the one at rest. The positive direction is for when the distance between them is decreasing.


Hence, The Doppler’s formulas for the two situations cannot be the same.


(iii) In case of the light waves, In vacuum, the above situations (a) and (b) are identical because the speed of light is a universal constant and does on depend on the motion of the observer or the source and thus the value always remains the same. But when light travels in a medium, according to the wave theory of light the speed depends on the nature of the medium.



Question 6.

In double - slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?


Answer:

Angular width is related to the slit spacing (d) as:


θ = λ/d


Thus,


d = λ/θ


Where,


λ is the wavelength of light


θ is the angular width of the fringe on the screen


d is the spacing between the two slits.


Given,


Wavelength of light, λ = 600 nm = 600 × 10 - 9 m


The angular width of the fringe θ = 0.1°


Note: The given angular width is in degrees, We must convert degrees into radiant for the angular width of the fringe.



Thus, Angular width, θ of the fringe can be written in radians as



⇒ 


d = 


⇒ d = 3.44 × 10 - 4 m


The spacing between the two slits “d” is 3.44 × 10 - 4 m



Question 7.

Answer the following question:

In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?


Answer:

If the width of the slit is made double the original width in a single slit diffraction experiment, the size of the central diffraction band will become half of the original value.


Since, we have the first minima occurs at


b sinθ = nλ


If we take n = 1, for the central maxima



Where,


θ is the angle between the wave ray at the point on the screen and the normal at the centre of the slit.


The space between the first minimums contains the central maxima.


If we increase the width will thus decrease.


The intensity depends on the amplitude as


I ∝ A2


Thus, when we double the size of the slit we double the amplitude and hence, the intensity becomes four times that of original intensity.


When central maxima are reduced to half, logically the same light has to be distributed in a smaller region. Increasing the intensity of central band maxima will make the intensity four times that of the original intensity.



Question 8.

Answer the following question:

In what way is diffraction from each slit related to the interference pattern in a double - slit experiment?


Answer:

If we close on of the slits in the case of a double slit experiment, the given apparatus now work for a single slit diffraction experiment. The central maxima and other secondary maxima from each individual single slits interfere to form a central maxima and other maxima for the double slit experiment due to their path differences when these slits work together as a coherent source.



Question 9.

Answer the following question:

When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?


Answer:

The diffraction pattern is observed when the size of the object is comparable to the wavelength of the diffracting light.


When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. This bright spot is a central maxima observed in the case of diffraction from the single slit. This is because light waves are diffracted from the edge of the tiny circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.



Question 10.

Answer the following question:

Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily?


Answer:

Bending the waves by obstacles at a large considerable angle only if the size of the obstacle is comparable to the wavelength of the waves.


If the wavelength of light waves is too small compared to the size of the obstacle, then the diffraction would not be considerable. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.



Question 11.

Answer the following question:

Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?


Answer:

The ray optics assumption is used in understanding location and several other properties of images in optical instruments because in the commonly used optical instruments are large when compared to the wavelength of light. Thus, the diffraction, a phenomenon is not observed when light waves interact on such a macro level and there is no violation of rectilinear property of light.



Question 12.

Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?


Answer:

Fresnel’s distance is given by the relation, Zp = a2 / λ


Where,


λ is the wavelength of the radio waves


Zp is the Fresnel distance, which shows that on distance smaller than this, the spreading due to diffraction smaller compared to the size of the beam.



If we want to prevent diffraction of the radio waves , According to the validity of ray optics, the distance of the middle should be less than The size of aperture “a” = 50 m


Z= 20 km = d/2 = 2 × 104 m


Aperture can be taken as:


a = 50 m


The longest wavelength of radio waves which can be sent without any hindrance due to diffraction effect,


λ = a2 / Zp


⇒ 


⇒ λ = 1250 × 10 - 4 m = 12.5 cm


Therefore, the longest wavelength of the radio waves is 12.5 cm



Question 13.

A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.


Answer:

The relation between fringe width “x” and the order of minima “n” is given as:



or



where,


d = width of the screen


D = distance between the screen and the slit


n = order of minima


λ = wavelength of the light


Given,


λ = 500 nm = 500 × 10 - 9 m


Distance between the screen and the slit , D = 1 m


For first minima, n = 1


x = 2.5 mm = 2.5 × 10 - 3 m


Distance between the slits = d


Distance of the first minimum from the centre of the screen can be obtained as:



⇒ d = 2 × 10 - 4 m = 0.2 mm


Therefore, the width of the slits is 0.2 mm.



Question 14.

Answer the following question:

When a low flying aircraft passes overhead; we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.


Answer:

Flying aircrafts transmits radar waves. When a low flying aircraft passes overhead, this radar signal however weak, can be received by the antenna and thus interferes with the TV signals. Thus, this interference causes a slight shaking of the picture that we sometimes notice on our TV screen.



Question 15.

Answer the following question:

As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns.

What is the justification of this principle?


Answer:

The linear superposition describes the linear character and solutions of the differential equation that governs the wave motion. The intensity distributions in both diffraction and interference patterns require the linear superposition to deduce and characterize the wave equation. If we take X1 and X2 as the displacement covered by the two waves. We can infer the resultant displacement of the two waves given as, X = X1 + X2



Question 16.

In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitably dividing the slit to bring out the cancellation.


Answer:

The path difference between the beams at each ends of the slits can be written as


bsinθ = nλ


where,


b is the size of the slit,


n is the order of the minima


θ is the phase angle.


The observations state that the intensity is zero at 


First, Let us divide the slit into two equal parts. Then a path a difference of λ/2 i.e. π is introduced between the waves of each part. The waves then superimpose to interfere destructively and thus producing a zero intensity minima on the screen. We can thus observe the same results if we divide the screen accordingly for n = 2,3,4….


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