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.

AP Physics B- Optics: Equations to be Remembered in Geometric Optics

The topics included under Optics in the AP Physics B Exam are the following

(a) Physical Optics: Interference and diffraction, Dispersion of light and the electromagnetic spectrum.

(b) Geometric optics: Reflection and refraction, Mirrors, Lenses.

These topics carry 10% of the total points (5% for Physical Optics and 5% for Geometric Optics )

You have to remember the following equations in Geometric optics:

(1) When a plane mirror is rotated through an angle θ, the reflected ray rotates through an angle 2θ.

(2) The minimum size (length) of a plane mirror for enabling a person of height ‘h’ to see his full image is h/2. You can easily prove this using the law of reflection (angle of incidence = angle of reflection).

(3) The radius of curvature (R) of a spherical mirror is related to its focal length (f) as

R = 2f

(4) The distance of object (s_o), distance of image (s_i) and the focal length (f) of a spherical mirror are related by the law of distances as

1/f = 1/s_o + 1/s_i

[The above equation is often written as 1/f = 1/u + 1/v where ‘u’ and ‘v’ stand for the distance of object and the distance of image respectively].

When you deal with reflection and refraction at spherical surfaces, you have to adopt a suitable sign convention. Let us quote from the post dated 22^nd August 2006 in entrancephysics.blogspot.com in this context:

“The sign convention adopted widely in Optics is the Cartesian convention. The ray incident on the curved surface is to be considered as proceeding in the positive X-direction and you have to measure all distances from the pole, which is supposed to be the origin.

Occasionally, you may be given a ray diagram in which the incident ray may be proceeding from right to left. To avoid confusion, imagine that the direction of the incident ray is still in the positive X-direction. You can even redraw the diagram to make the incident ray proceed from left to right if you want. The signs given to the distances are as in the Cartesian coordinate system: pole to right positive and pole to left negative. Distances measured upwards are positive and those measured downwards are negative but you will mostly encounter problems with leftward and rightward measurements.

While solving problems, you should apply the signs to all known quantities. The unknown quantities are left as they are in the formulae. You will be able to arrive at conclusions by interpreting the sign of the unknown quantity you finally arrive at as the answer. For instance, if the distance of an image is obtained as negative, you will immediately understand that the image is on the same side of the curved surface as the object is”.

According to the Cartesian sign convention, the radius of curvature (R) and the focal length (f) of a concave mirror are negative since you have to measure these distances leftwards (opposite to the direction of the incident ray), from the pole P (fig.).

In the case of a convex mirror, R and f are positive since you have to measure these distances rightwards (in the same direction as that of the incident ray), from the pole P. Distance of a real object will be always negative since it is to be measured from pole to the object (towards the left and opposite to the direction of the incident light).

You have to use the Cartesian sign convention in problems involving reflection as well as refraction at spherical surfaces. But, don’t be scared of the Cartesian sign convention. It’s simple. You will get accustomed to this sign convention when we discus problems involving spherical surfaces.

(5) Magnification (M) produced by a mirror is the ratio of the size of image to the size of object and is given by

M = h_i/h_o = –s_i/s_o

When we apply the sign convention, for the upright (and virtual) image formed by a convex or concave mirror, the magnification is positive. If the image is inverted (and real), the magnification is negative.

(6) n₁ sin θ₁ = n₂ sin θ₂ where n₁ is the refractive index of medium 1, n₂ is the refractive index of medium 2, θ₁ is the angle of incidence (in medium 1) and θ₂ is the angle of refraction (in medium 2).

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

n = n₂/ n₁ = sin θ₁/ sin θ₂

If medium1 is air or free space, the refractive index of medium 2 is given by

n₂ = n = sin θ₁/ sin θ₂ = sin i/sin r, on putting θ₁ = i and θ₂ = r

Note that when you say ‘refractive index of a medium’ without mentioning a reference nedium, you mean the refractive index with respect to free space (or air). The refractive index of free space (or air) is unity. [When air is compressed, its refractive index is increased]

[Refractive index(n) of medium 2 with respect to medium 1 is also 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].

(7) When an object in medium 2 is viewed from medium 1, normally through the plane surface separating the two media, the apparent distance of the object (which is the distance of the refracted image) from the surface is related to the real distance as

Apparent distance = (Real distance)/n

where ‘n’ is the refractive index of medium 2 with respect to medium 1.

In the figure, O is the object and I is the image formed by normal refraction at the plane surface AB separating the two media.. PI is the apparent distance and PO is the real distance of the object O.

The apparent depth (d_app) of a liquid in a container when viewed normally from above is given by

d_app = d_real /n

where d_real is the real depth and ‘n’ is the refractive index of the liquid.

If there are many layers of liquids of refractive indices n₁, n₂, n₃,….etc, with real thickness t₁, t₂, t₃ …etc, the apparent thickness (t_app) of the entire liquid column is given by

t_app = t₁/n₁ + t₂/n₂ + t₃/n₃ +….etc

Evidently, the above equation is true in the case of parallel slabs of transparent media piled up to make a composite slab.

(8) sin θ_c = n₂/n₁ where θ_c is the critical angle (for total internal reflection) of the medium of higher refractive index n₁ with respect to the medium of lower refractive index n₂.

If light is proceeding from a (denser) medium of refractive index ‘n’ to air or free space, the critical angle of the medium is given by

sin θ_c = 1/n

(9) The angle of minimum deviation ‘D’ produced by a prism is related to the refractive index ‘n’ of the material of the prism as

n = sin[(A+D)/2] / sin(A/2) where A is the angle of the prism.

(10) When refraction occurs at a spherical surface of radius of curvature R, separating two media of refractive indices n₁ and n₂, the object distance s₀ and the image distance s_i are related to R, n₁ and n₂ by the equation,

n₂/s_i – n₁/s_o = (n₂ – n₁) /R

[This is often written as n₂/v – n₁/u = (n₂ – n₁) /R, using symbols ‘v’ and ‘u’ for the image distance and the object distance respectively].

(11) Lens maker’s equation is

1/f = [(n₂/n₁) – 1] (1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, n₂ is the refractive index of the material of the lens and R₁ and R₂ are the radii of curvature of its surfaces. n₁ is the refractive index of the medium in which the lens is placed.

Note that if the lens is placed in free space (or air at ordinary pressures), the refractive index n₁ = 1. Putting n₂ = n, the lens maker’s equation becomes

1/f = (n– 1) (1/R₁ – 1/R₂)

In the case of a biconvex lens, the sign of R₁ will be positive and the sign of R₂ will be negative so that the above equation will become

1/f = [(n₂/n₁) – 1] (1/R₁ + 1/R₂)

In the case of a biconcave lens, the sign of R₁ will be negative and the sign of R₂ will be positive so that the above equation will become

1/f = [(n₂/n₁) – 1] (–1/R₁ – 1/R₂) or,

1/f = – [(n₂/n₁) – 1] (1/R₁ + 1/R₂)

The focal length of a convex lens is thus positive while that of a concave lens is negative [if the medium in which the lens is placed is rarer (n₁<>2)].

(12) Law of distances for a lens is

1/f = 1/s_i – 1/s_o

[This equation is often written as 1/f = 1/v – 1/u where the distance of object s_o and the distance of image s_i in the above equation are denoted by the symbols ‘u’ and ‘v’ respectively].

In the case of a converging lens forming a real image, on applying the signs in accordance with the Cartesian convention, the above equation will become

1/f = 1/s_i + 1/s_o since the sign of s_o is negative and the signs of s_i and f are positive.

(13) Magnification (M) produced by a lens is the ratio of the size of image to the size of object and is given by

M = h_i/h_o = s_i/s_o

When we apply the sign convention, for the upright (and virtual) image formed by a converging or diverging lens, the magnification is positive. If the image is inverted (and real), the magnification is negative.

(14) Focal power (or simply, power) of a lens is the reciprocal of focal length in metre. Therefore, power (P) is given by

P = 1/f

Unit of focal power is dioptre. The sign of ‘f’ is to be considered when you express the power. Therefore, converging lenses have positive power while diverging lenses have negative power.

(15) The effective focal length (F) of two or more thin lenses (of focal lengths f₁, f₂, f₃, …etc.) in contact is given by the reciprocal relation

1/F = 1/f₁ + 1/f₂ + 1/f₃ + …..etc.

This is really a power equation which says that the effective power is the (algebraic) sum of individual powers.

You can obtain informations such as the position of the image, nature of the image etc., by resorting to the method of ray tracing in many situations. The important things you must remember in this context are:

(i) A ray proceeding parallel to the principal axis will pass through the principal focus in the case convex lenses (to be more general, converging lenses) and concave mirrors. In the case of concave lenses (diverging lenses) and convex mirrors, the ray will appear to diverge from the principal focus.

(ii) A ray that passes through or appears to pass through the centre of curvature of a mirror will retrace the path after reflection at the mirror. A ray passing through the optic centre of a lens (converging as well as diverging) will proceed without deviation.

(iii) The law of reflection which states that the angle of incidence is equal to the angle of reflection is true for all surfaces (plane as well as spherical).

In the next post we will consider questions on geometric optics. Solution of problems based on the method of ray tracing as well by the application of the formulas mentioned above will be discussed.