AP Physics B- Free Response Question (for practice) on Interference

Let us discuss a free-response question of the type you can expect in the section ‘interference’.

At the moment I just give you the question for you to try yourself. (I’ll give you the answer in the next post). Here is the question:

Your teacher suggests to you that an interference method could be used to measure the frequency of the sound emitted by two small identical speakers S₁ and S₂ arranged horizontally with a separation of 0.8 m (or any suitable distance) as shown. The speakers are connected in series to a sine wave generator of constant output frequency. Your teacher moves the probe (microphone) of a sound level meter along the horizontal line AB parallel to the line S₁ S₂ and demonstrates to you that at O, equidistant from S₁ and S₂, the intensity of sound is maximum and it varies between maximum and minimum when the probe is moved along OA and OB.

(a) Explain briefly what measurements you will make to enable you to calculate the frequency of the sound emitted by the speakers.

(b) If the distance between the speakers is increased, what quantity you measure is affected and how?

(c) What equations you will use for calculating the frequency? Explain the reason for your answer to (b) based on the relevant equation.

(d) If the distance between the 2^nd maximum on one side of O and the 3^rd maximum on the other side of O is 1.725 m, calculate the frequency of the sound emitted by the speakers, assuming that the speed of sound at the laboratory temperature is 345 ms^–1.

(e) Your teacher asks you to reverse the connection to the terminals of one of the speakers. What change, if any, will this produce in the location of the maxima and minima? Give reason for your answer.

Try to answer the above question which carries 15 points. I’ll be back with the answer shortly.

AP Physics B- Multiple Choice Questions on Interference and Diffraction

In the post dated 6^th January 2008, the essentials of wave optics required for you to work out multiple choice questions were discussed. (You will find that post by clicking on the label ‘wave optics’ below this post). In that post, the following question was given for your practice:

A student prepared a Young’s double slit by drawing two parallel lines (with a separation less than a millimeter) on a smoked glass plate. [For smoking, the plate was held above the flame of a kerosene lamp]. Accidentally he used two different pins to draw the lines so that the widths of the slits (regions from where smoke was removed by the pin) were in the ratio 4:1. What will be the ratio of the intensity at the interference maximum to that at the interference minimum?

(a) 4

(b) 2

(c) 3

(d) 6

(e) 9

As promised, I give the solution here:

The ratio of intensities of light proceeding from the slits, I₁: I₂ = 4 : 1

Therefore, the ratio of amplitudes of light proceeding from the slits,

a₁:a₂ = √I₁: √I₂ = 2:1

The amplitudes at the maximum and minimum of the interference pattern are therefore in the ratio (a₁+a₂) : (a₁–a₂) = 3:1

The intensities at the maximum and the minimum are in the ratio3²:1² = 9:1 [Option (e)].

[The intensity is proportional to the square of the amplitude].

We will discuss a few more typical multiple choice questions in this section:

(1) A single slit diffraction pattern is obtained on a screen using yellow light. If the yellow light is replaced by blue light without making any other changes in the experimental set up, what will happen to the diffraction bands?

(a) Bands will disappear

(b) Bands will become broader and farther apart

(c) Bands will become broader and crowded together

(d) Bands will become narrower and farther apart

(e) Bands will become narrower and crowded together

The angular width of the central maximum is 2λ/a and the minima are located at θ = ± λ/a, ± 2λ/a, ± 3λ/a…etc.

This shows that when the wave length is decreased as is the case when blue light is used instead of yellow light, the bands will become narrower and crowded together [Option (e)].

(2) If white light is used in Young’s double slit experiment, what will happen to the interference bands?

(a) No bands will be obtained

(b) Many bands will be obtained as in the case of monochromatic light, but they will be coloured except the centre of the central band which will be white

(c) Very few bands will be obtained, but they will be coloured except the centre of the central band which will be white

(d) Many bands will be obtained as in the case of monochromatic light, but all of them will be white

(e) Very few bands will be obtained, but all of them will be white.

The centre of the central bright band (n = 0) will be white since all colours will satisfy the condition of zero path difference (and therefore of brightness). Since the condition for the first (order n =1) bright band is satisfied, at the point nearest to the centre, by the shortest wave length (violet), the first bright band will be coloured with the edge nearest to the centre violet and the edge farthest to the centre red. The remaining bright bands also will be coloured likewise. There will be very few distinct bands since the same point on the screen will satisfy the condition for brightness for one wave length and the condition for darkness for another.

The correct option therefore is (c).

(3) In an arrangement for Young’s double slit experiment, the separation between the slits is 1 mm. It is found that 8 bands of the double slit interference pattern can occupy the central maximum of the single slit diffraction pattern produced by one of the slits. What is the wjdth of each slit?

(a) 0.2 mm

(b) 0.25 mm

(c) 0.3 mm

(d) 0.35 mm

(e) 0.4 mm

The angular fringe width in the case of the double slit interference pattern is λ/d so that the total angular separation of 8 interference bands is 8λ/d. Since the angular width of the central maximum in the case of the single slit diffraction pattern is 2λ/a, we have

8λ/d = 2λ/a.

Putting d = 1 mm, a = d/4 = 0.25 mm.

[You should note that in Young’s double slit experiment, the pattern you see on the screen is actually a superposition of single-slit diffraction from each slit, and the double-slit interference pattern. Therefore, there will be a broader diffraction peak in which several fringes of smaller width due to double-slit interference appear. The angular width of a double slit intetrference fringe is λ/d where as the angular width of the central maximum due to the single slit diffraction is 2λ/a. Therefore, the number of interference fringes occurring in the broad central diffraction maximum depends on the ratio d/a, that is the ratio of the distance between the two slits to the width of a slit. If ‘a’ is very small, the diffraction pattern will become very flat and we will observe the two-slit interference pattern with an appreciable number of interference maxima of almost equal intensity in the central region].

You will find some useful multiple choice questions in this section at physicsplus: Multiple Choice Questions on Wave Optics.

AP Physics B- Optics: Equations to be Remembered in Interference and Diffraction

If I have seen a little further it is by standing on the shoulders of Giants.

–Sir Isaac Newton

In Physical Optics the topics included are: (i) Interference and diffraction (ii) Dispersion of light and the electromagnetic spectrum. These topics carry 5% of the total points.

You must remember the following equations to make you strong in answering multiple choice questions involving interference and diffraction:

(1) Refractive index(n) of medium 2 with respect to medium 1 is given given by

n = n₂/ n₁ = λ₁/ λ₂ = v₁/v₂ where λ₁ and λ₂ are the wave lengths and v₁ and v₂ are the velocities of light in medium 1 and medium 2 respectively].

(2) The resultant displacement produced (at a particular point in a medium) by a number of waves is the vector sum of the displacements produced by each of the waves. For instance, if two waves arriving at a point P produce displacements y₁ and y₂, the resultant displacement at P is the vector sum of y₁ and y₂. If the interfering waves have amplitudes a₁ and a₂ and they arrive at the point P with a phase difference of φ, the resultant amplitude at P will be [a₁²+ a₂²+ 2a₁ a₂ cosφ]^1/2, in accordance with vector addition. If the interfering waves have intensities I₁ and I₂ the resultant intensity (I) at P will be given by

I = I₁ + I₂ + [2√(I₁I₂)] cosφ, since the intensity is directly proportional to the square of the amplitude.

If φ = 0, 2π, 4π, 6π,…..(an even multiple of π), the intensity is maximum (I_max) given by

I_max= I₁ + I₂ + [2√(I₁I₂)]

If φ = π, 3π, 5π, 7π,…..(an odd multiple of π),, the intensity is minimum (I_min) given by

I_min= I₁ + I₂ – [2√(I₁I₂)]

(3) Two sources of light (or any wave) are said to be coherent if they produce waves of the same frequency with a constant phase difference between them. (The condition of same frequency is implied in the condition of constant phase difference; yet it is usually specified when we define coherent sources). If two coherent sources produce waves of equal amplitude (a) and therefore equal intensity (I₀), then the resultant intensity (I) when they interfere is given by

I = I₀ + I₀ + [2√(I₀I₀)]cosφ = 2I₀(1+cosφ)

If the phase difference between the interfering waves (of equal intensity) is zero or an even multiple of π (in other words, if the path difference between the waves is zero or an integral multiple of the wave length λ), the intensity is maximum (I_max) given by

I_max = 2I₀(1+1) = 4I₀ …..(Path difference = nλ where n = 0,1,2,3,4,…..etc)

If the phase difference between the interfering waves (of equal intensity) is an odd multiple of π (in otherwords, if the path difference between the waves is an odd multiple of half wave length), the intensity (I_min) is zero given by

I_min = 2I₀(1–1) = 0……(Path difference = (2n+1)λ/2 where n = 0,1,2,3,4,….etc)

(4) In the interference fringe pattern produced by Young’s double slit, the distance of the n^th bright fringe (x_n) from the central (zero order) bright fringe is given by

x_n = nλD/d, where n = 0, ±1, ±2, ±3….etc.

where λ is the wave length of light, D is the distance between the screen (on which the fringes are formed) and the double slit and ‘d’ is the separation between the two slits S₁ and S₂ forming the double slit (fig.).

The distance of the n^th dark fringe (x_n) from the central (zero order) bright fringe is given by

x_n = (2n+ 1) λD/2d, where n = 0, ±1, ±2, ±3….etc. [We use ± sign for the integers to indicate the fringes on both sides of the central fringe]

The fringe width β is the distance between the centres of consecutive brigt fringes or dark fringes and is given by

β = x_n+1 –x_n

Therefore, β = λD/d

Angular fringe width is independent of the distance of the screen and is given by

β_θ = β/D = λ/d

(5) When light traveling through a rarer medium gets reflected from the surface of a denser medium, it suffers a phase change of π, which is equivalent to a path change (path difrference) of λ/2

(6) Colours in thin transparent films can be observed in reflected light as well as transmitted light. The condition for brightness in the reflected light is

2μtcosr = (2n+ 1)λ/2 where μ is the refractive index of the film, t is its thickness, r is the angle of refraction in the film and n = 0,1,2,3,4,….etc.

[A film of negligible thickness cannot satisfy the above condition and hence it cannot appear bright in reflected light. (It will appear dark)].

The condition for brightness in transmitted light is

2μtcosr = nλ where n = 0,1,2,3,4,….etc.

[A film of negligible thickness can satisfy the above condition (for n = 0) and hence it will appear bright in transmitted light].

You should note that Newton’s rings in reflected light are produced by the interference between the light waves reflected from the two surfaces of an air film contained in between a glass plate and a lens.

(7) In the diffraction pattern produced by a single slit, the angular separation between the first minima is 2λ/a where a is the width of the slit. The angular width of the central maximum is therefore 2λ/a.

The centre of the central maximum is at O (fig.). On either side of O, positions of minimum intensity are obtained when the angle θ = ±nλ/a where n = 1,2,3…etc.

Therefore, the minima are located at θ = ± λ/a, ± 2λ/a, ± 3λ/a…etc. (Fig.) Positions of

secondary maxima are obtained on either side of O corresponding to

θ = ± (2n+ 1)λ/2a where n = 1,2,3….etc.

Therefore, the secondary maxima are located at θ = ±3λ/2a, ±5λ/2a, ±7λ/2a….etc. The intensity of the secondary maxima goes on decreasing. [The first secondary maximum has intensity less than 5% of the intensity of the central maximum].

(8) In a plane diffraction grating, the n^th principal maximum is obtained at an angle θ given by

(a+b) sinθ = nλ where a is the width of the slit and b is the width of the opaque region between the slits. [a+b is called the grating element]

In the next post questions on wave optics will be discussed. Here is a question which you must be able to work out in no time if you have understood the phenomenon of interference.

A student prepared a Young’s double slit by drawing two parallel lines (with a separation less than a millimeter) on a smoked glass plate. [For smoking, the plate was held above the flame of a kerosene lamp]. Accidentally he used two different pins to draw the lines so that the widths of the slits (regions from where smoke was removed by the pin) were in the ratio 4:1. What will be the ratio of the intensity at the interference maximum to that at the interference minimum?

(a) 4

(b) 2

(c) 3

(d) 6

(e) 9

Work out the above question. I’ll be back shortly with the solution and of course, more solved questions.

AP Physics B- More Questions (MCQ) from Geometric Optics

We will discuss few more questions on Geometric Optics.

Consider the following MCQ on normal refraction:

Suppose a glass slab of thickness ‘t’ and refractive index ‘n’ is placed on a printed page of a book you are reading. The prints will

(a) appear to be shifted towards you by (t – ^t/_n)

(b) appear to be shifted towards you by ^t/_n

(c) appear to be shifted away from you by (t – ^t/_n)

(d) appear to be shifted away from you by ^t/_n

(e) not appear to be shifted

The prints will be shifted towards you since you are viewing the prints through a denser medium. The apparent thickness of the glass slab will be t/n where as its real thickness is t. The print in contact with the glass slab will appear to be raised (shifted towards you) by the difference (t – ^t/_n) between the real thickness and the apparent thickness

You should note that an object placed at a distance equal to 2f from a converging lens or a concave mirror of focal length f will result in a real image at distance 2f. Consider the following MCQ which is meant to check your awareness of this as well as the apparent shift produced by a transparent slab:

A glowing light emitting diode (LED) is placed on the principal axis of a convex lens at a distance of 24 cm from it. The focal length of the lens is 10 cm. When a glass slab of thickness ‘t’ and refractive index 1.5 is interposed between the lens and the LED, a well defined real image of the LED is formed at 20 cm on the other side of the lens. What is the thickness of the glass slab?

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 12 cm

(e) 16 cm

Since the real image is formed by the lens at a distance 2f (= 20cm), the object must appear for the lens to be located at a distance of 20 cm as shown in the figure in which O is the object, I is the image formed by normal refraction at the slab and I₁ is the final real image produced by the lens. In other words, the glass slab must produce an apparent shift of (24–20) = 4 cm so that the image I formed by normal refraction serves as the object for the lens. [The glass slab can be placed anywhere between the lens and the LED to produce this shift]. Therefore we have

(t – ^t/_n) = 4cm.

Substituting for n = 1.5, we have

(1.5t – t)/1.5 = 4 cm, from which t = 12 cm.

Solution of the following question demands just the basic knowledge in geometry and geometric optics:

Light is incident normally on face BC of a glass prism ABC as shown. A liquid of refractive index ‘n’ is placed on the face AB of the prism. If the refractive index of glass is 3/2, the maximum value of n for which total internal reflection at the face AB will occur is

(a) √3

(b) √2

(c) 2/√3

(d) (3√3)/2

(e) (3√3)/4

The angle of incidence of the ray of light on the face AB is 60º (noting that the angle of incidence is the angle between the ray and the normal to AB at the point of incidence).When the refractive index of the liquid increases, the critical angle for total reflection at AB also increases in accordance with the equation

sin θ_c = 1/n_' where n' is the refractive index of glass with respect to the liquid given by

_n' = (Refractive index of glass)/ (Refractive index of liquid) = (3/2)/n = 3/2n.

Since the angle of incidence of 60º is to be the critical angle, the maximum possible value of n is given by

Sin 60º = 1/(3/2n) = 2n/3.

Or, √3/2 = 2n/3, from which n = (3√3)/4

You should note that a converging lens will become diverging (and a diverging lens will become converging) if the lens is immersed in a medium of refractive index greater than that of the material of the lens. Let us discuss a problem to high light this point:

A thin biconcave lens made of glass of refractive index 1.5 has focal length 20 cm in air. In a liquid of refractive index 1.6 it will behave as

(a) diverging lens of focal length 60 cm

(b) diverging lens of focal length 120 cm

(c) converging lens of focal length 60 cm

(d) converging lens of focal length 120 cm

(e) converging lens of focal length 160 cm

We have the lens maker’s equation,

1/f = [(n₂/n₁) – 1] (1/R₁ – 1/R₂) with usual notations.

A concave lens in air is diverging and f is negative, R₁ is negative and R₂ is positive in accordance with the Cartesian sign convention. Therefore we have

1/(–20) = [(1.5/1) – 1] (–1/R₁ – 1/R₂), when the lens is in air.

[We don’t know the values of R₁ and R₂. These are not required in this question. But you must note that in the case of equi-convex and equi-concave lenses of refractive index 1.5, the radius of curvature will be equal in value to the focal length. You can check this using lens maker’s equation].

When the lens is within the liquid, we have

1/f = [(1.5/1.6) – 1] (–1/R₁ – 1/R₂)

[We don’t know the sign of the focal length f. It will be obtained as positive or negative depnding on whether the lens is converging or diverging in the liquid].

From the above two equations (on dividing one by the other), we obtain

–f /20 = (0.5×1.6)/(–0.1), from which f = 160 cm.

The sign of ‘f’ is positive and hence the lens will behave as a converging lens of focal length 160 cm.

You will find some useful multiple choice questions at physicsplus

AP Physics B- Multiple Choice Questions from Geometric Optics

The following multiple choice questions (MCQ) will help to boost your morale in facing the AP Physics B Exam:

(1) At what angle with respect to the horizontal, does a fish under water see the setting sun?

(a) tan^–1(^√7/₃) (b) tan^–1(⁵/₃) (c) tan^–1(⁵/_√7)

(d) tan^–1(^√5/₃) (e) tan^–1(^√7/₄) 

With reference to the figure, S represents the setting sun from which the ray SAF reaches the fish F after refraction at F. For the fish the ray appears to come from S₁ which is the virtual image of the sun. Evidently angle FAN is the critical angle (θ_c) for the water-air interface. Since n = 1/sinθ_c, we have

sinθ_c = 1/n = 1/(4/3) =3/4

[Right angled triangle FAB is drawn side by side with the figure showing the refraction, to show the opposite side of length 3 units and adjacent side of length 4 units. The hypotenuse of this triangle is √7].

The rays coming from the sun make an angle S₁AS = FAB with the horizontal. The angle required in the problem is angle FAB, which is tan^–1(^√7/₃).

(2) In the adjoining figure, SS is a spherical surface separating two media of refractive indices n₁ and n₂ where n₁ > n₂. C is the centre of curvature of the spherical surface. An observer, keeping his eye beyond C in the medium of refractive index n₂ views the refracted image of an object AB placed as shown in the medium of refractive index n₁. The image will be

(a) real, inverted and magnified

(b) real, upright and diminished

(c) virtual, upright and diminished

(d) virtual, upright and magnified

(e) virtual, inverted and diminished

You can easily solve this problem by ray tracing as shown in the figure.

CN is normal to the spherical surface. A ray BC passing through the centre of curvature C does not suffer any deviation. Another ray BD is deviated at D obeying Snell’s law of refraction, n₁ sin θ₁ = n₂ sin θ₂, where θ₁ is the angle of incidence BDN and θ₂ is the angle of refraction EDC. In this question you are asked to find the nature of the image qualitatively and you need not worry about the actual values of θ₁ and θ₂. But you must be careful to draw the refracted ray DE so as to have θ₂ larger than θ₁ since n₁ > n₂.

The diverging ray DE produced backwards meets the ray BC at M and IM is the virtual, upright and diminished image of the object AB. So, option (c) is correct.

Suppose you were asked to calculate the distance of the image and its magnification. In that case, the refractive index values, the radius of curvature of the spherical surface and the distance of the object will be given in the problem. You will have to work out things quantitatively. If you still want to resort to the method of ray tracing, you should draw all distances to a suitable scale. The spherical surface SS is to be first drawn by fitting the radius to the chosen scale. The object AB is to be shown at the given distance (to the scale). After drawing the ray BD you will have to measure the angle of incidence θ₁ ( angle BAN) and then calculate angle of refraction θ₂ (angle EDC) using n₁ sin θ₁ = n₂ sin θ₂. The refracted ray DE is to be drawn so that angle EDC is equal to the calculated value of θ₂. After completing the ray diagram, you will have to measure the distance of the image (PI). The size of the image (IM) need not be measured since you can calculate magnification (M) using the equation M = s_i/s_o.

If a question does not demand you to do things specifically by ray tracing, you can calculate all quantities using the equations we considered in the post dated 30^th December 2007.

Let us modify the above problem to make things quantitative:

In the adjoining figure, SS is a spherical surface separating a transparent medium of refractive index 4/3 from air. C is the centre of curvature of the spherical surface. An observer located in air views the refracted image of an object AB of height 10 cm placed as shown in the medium of refractive index 4/3. If the distance of the object (from the spherical surface) is 3m, calculate the size of the image.

We have n₂/s_i – n₁/s_o = (n₂ – n₁) /R.

Here n₁ = 4/3, n₂ = 1 (air), s_o = –3m, R = 2m.

Therefore, 1/s_i – (4/3)/(–3) = [1– (4/3)] /2

The sign of the object distance s_o is negative since we have to measure if from P to A, opposite to the direction of the incident ray. The sign of R is positive since we have to measure it from P to C in the same direction as that of the incident ray.

The above equation gives s_i = –18/11 m.

The magnification is M = s_i/s_o = (–18/11)/ (– 3) = 18/33 = 0.545

Since the magnification is positive, the image is upright and virtual. We can obtain this information the moment we find that the sign of the image distance (s_i) is negative. [However, in the case of an image formed by reflection, the image distance will be negative for real and inverted image].

The size of the image = M× size of object = 0.545×10 cm = 5.45 cm.

(3) A straight rod of length very large compared to the focal length ‘f’ of a concave mirror lies along the principal axis of the mirror. The nearer end of the rod is at a distance ‘x’ from the pole of the mirror. If x > f, the length of the image of the rod is

(a) xf/(x– f)

(b) xf/(x+ f)

(c) f²/(x+f)

(d) f²/x

(e) f²/(x–f)

For a spherical mirror we have 1/f = 1/s_o + 1/s_i.

One end of the rod is far away from the mirror and hence the image of that end must be formed at the focus . [From the above equation, on applying the negative sign for f, you will get s_i = –f, the negative sign indicating that the image of the end is real and on the object side of the mirror].

The image of the near end of the rod will be formed at s_i given by

1/f = 1/x + 1/s_i so that

1/s_i = 1/f – 1/x from which s_i = xf/(x –f)

[If you apply the negative signs for f and x, in accordance with the Cartesian sign convention, you will get s_i = –xf/(x –f), the negative sign indicating that the image of the near end of the rod also is real and on the object side of the mirror].

The length of the image is [xf/(x –f)] – f = f²/(x–f)

We will discuss more questions on geometric optics in the next post.