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.

Answer to AP Physics Free Response Question on Inductance

A free response question involving inductance was posted on 23^rd December 2007 for your practice. This was the question:

In the circuit shown in the figure, the battery and the inductor have negligible resistance. The currents through R₁, R₂ and R₃ are i₁, i₂ and i₃ respectively. The switch S is closed at t = 0.

(a) Calculate the currents i₁, i₂ and i₃ immediately after closing the switch S (when t = 0).

(b) Calculate the currents i₁, i₂ and i₃ when steady state is reached.

(c) Show the nature of variation of the current through the battery graphically, with time ‘t’ on the X-axis and the current (i₁) on the Y-axis.

(d) What are the values of the slopes of the above graph at t = 0 and at t = ∞.

(e) The switch S is opened and the resistance R₃ is disconnected. If the switch S is now closed, what time is required for the current to attain 63.2% of the final steady value?

As promised, I give below the answer:

(a) Since the parallel branch containing R₃ does not contain any inductance, the current through that branch (i₃) rises to the maximum possible value in this condition immediately on closing the switch S. The same current now flows through R₁ since the initial current i₂ ( at t = 0) through the branch containing the inductance L and the resistance R₂ is zero. [Note that in an LR circuit, the current I at the instant ‘t’ is given by I = I₀(1– e^–Rt/L) where I₀ is the final steady current]

Therefore, when t = 0, i₁ = i₃ = (12 V)/ (8 Ω + 8 Ω ) = 0.75 A and i₂ = 0

(b) Steady state is reached when t = ∞. In this condition, the current is controlled by the resistances only [since dI/dt is zero and no opposing induced voltage L(dI/dt) is developed across the inductance]. The circuit then behaves as made of the parallel combination of R₂ and R₃ (having effective value of 4 Ω) in series with the battery and the resistance R₁. The effective resistance of the circuit is (8 + 4) Ω = 12 Ω and the current delivered by the battery (i₁) is (12 V)/ (12 Ω) = 1 A.

Since this current is equally divided between the two parallel branches, current through each branch is 0.5 A. So, i₂ = i₃ = 0.5 A.

(c) The variation of the current i₁ with time t is shown in the adjoining graph. [The current grows exponentially in the inductive branch and this is why the current delivered by the battery finally rises to the steady value of 1 A as shown].

(d) The slope of the graph when t = o is infinity since the current grows abruptly to 0.75 A because of the purely resistive parallel branch containing R₃. The slope when t = ∞ is zero since the current has settled to the final steady value.

(e) When R₃ is disconnected, We have a series LR circuit with L = 2 H and R = R₁ + R₂ = 16 Ω. The time constant of the circuit is L/R = 2/16 = 0.125 s. The current grows to 63.2 % of the final steady current within this time.

[Time constant is L/R which is the time required for the current to become (1 – 1/e) times the final steady current. This is obtained by putting t = L/R in the equation,

I = I₀(1– e^–Rt/L)].

Merry Christmas!

AP Physics C-Free Response Question on Inductance (for Practice)

In the circuit shown in the figure, the battery and the inductor have negligible resistance. The currents through R₁, R₂ and R₃ are i₁, i₂ and i₃ respectively. The switch S is closed at t = 0.

(a) Calculate the currents i₁, i₂ and i₃ immediately after closing the switch S(when t = 0).

(b) Calculate the currents i₁, i₂ and i₃ when steady state is reached.

(c) Show the nature of variation of the current through the battery graphically, with time ‘t’ on the X-axis and the current (i₁) on the Y-axis.

(d) What are the values of the slopes of the above graph at t = 0 and at t = ∞.

(e) The switch S is opened and the resistance R₃ is disconnected. If the switch S is now closed, what time is required for the current to attain 63.2% of the final steady value?

The above question carries 15 points. The division of points among the sections (a), (b), (c), (d) and (e) can be as 4 + 3 + 4 +2 +2. Try to answer the above question within 15 minutes or less. I’ll be back with the answer shortly.

AP Physics C - Multiple Choice Questions on Inductance

1. An air cored coil of self inductance L has N turns of fine insulated copper wire wound on a former of cross section area A. If the area and number of turns are doubled and the core is a medium of relative permeability 1000, the self inductance of the coil will be

(a) 8000 L (b) 4000 L (c) 8×10^–3 L (d) 4×10^–3 L (e) L

Self inductance of a coil is directly proportional to the area of cross section, relative permeability of the core and the square of the number of turns. The answer therefore is 8000 L.

2. A straight air cored solenoid has length 1 m, area of cross section 10 cm² and total number of turns 2000. If a current of 1 A flowing in it is reversed in 0.1 s, the average emf induced in it will be nearly

(a) 100V (b) 10 V (c) 1 V (d) 0.5 V (e) 0.1 V

The emf induced is given by

ε = –L(dI/dt),

where dI is the change in current in the time dt and L = μ₀n²Aℓ where ‘ℓ’ is the length of the solenoid, ‘A’ is its cross section area, ‘n’ is the number of turns per metre and μ₀ is the permeability of air (or free space) which you must remember as 4π×10^–7.

Ignoring the negative sign which is because of Lenz’s law, we have

ε = μ₀n²Aℓ (dI/dt)

Substituting, ε = 4π×10^–7×(2000)²×(10×10^–4)×1×[1– (–1)]/(0.1) volt.

Note that the change in the current is 2 A ( from 1 A to –1 A) since the current is reversed.

Therefore, ε = 0.1 (nearly).

3. In the circuit shown, the switch S is kept closed so that the current flowing in the circuit is the final steady value. When the switch S is opened, the time constant for the decay of current in the inductance is

(a) L/R

(b) L/5R

(c) 5L/4R

(d) 4L/5R

(e) L/3R

The current decays through the resistors R and 4R in series with L and therefore the time constant is L/(R+4R) = L/5R.

4. In the above question, suppose the resistance 4R is disconnected and is then connected between the inductance L and the negative terminal of the battery. The switch S is kept closed. If the switch is opened, the time constant for the decay of current in L is

(a) L/R

(b) infinity

(c) zero

(d) L/5R

(e) 5L/4R

This is a simple case which may however generate confusion in your mind. Note that when you open the switch, the resistance in series with L is infinite and the time constant will be L/∞ = 0.

Physically, the current becomes zero abruptly since there is no conducting path for the current to flow.

[Discharging of a charged capacitor through a resistor is an easy thing since the capacitor can retain the charge during the small time during which the capacitor is disconnected from the charging battery and then connected to the resistor, using a charge-discharge key. In the case of the LR circuit, for studying the current decay pattern, after establishing the steady final current through the circuit, you have to short circuit the leads connected to the battery and immediately disconnect the battery so that the battery is not damaged].

5. A small plane circular coil of radius ‘r’ having ‘n’ turns is placed at a distance ‘d’ (d >> r) from a straight vertical conductor. The coil and the straight conductor are contained in the same vertical plane. The mutual inductance between the coil and the straight conductor is

(a) μ₀ nr²/2πd

(b) π μ₀ nr²/2d

(c) μ₀ nr²/2

(d) μ₀ nr²/2π²d

(e) μ₀ nr²/2d

The mutual inductance is given by

M = Ф/I where I is the current in the straight conductor and Ф is the magnetic flux through the coil.

Since the coil is small, the magnetic field produced at the coil by the current in the straight conductor can be assumed to be constant over the entire area of the coil so that

Ф =nBA = n(μ₀I/2πd)(πr²)

Therefore, M = μ₀ nr²/2d

You will find similar multiple choice questions at physicsplus: Multiple Choice Questions (MCQ) involving Inductance

AP Physics C: Inductance- Equations to be remembered

The following equations are to be remembered in this section:

1. Self inductance (coefficient of self induction) ‘L’ of a circuit is defined in terms of magnetic flux linkage as

L = Ф/I where Ф is the magnetic flux (in weber) linked with the circuit when a current I ampere flows in the circuit.

L is defined in terms of the self induced voltage ‘ε’ as

L =ε /(dI/dt) where dI/dt is the rate of change of current in the circuit.

The induced voltage opposes the change in the current in the circuit in accordance with Lenz’s law and hence the above equation has to be strictly

L = – ε/(dI/dt). But remember that self inductance is a positive quantity. It is the electromagnetic analogue of mass (which has the property of inertia) in mechanics.

[Induced voltage (emf) is represented by the symbol V also, instead of ε].

It will be better to remember the above equation as

ε = –L(dI/dt).

2. An inductance L carrying a current I possesses energy U_L given by

U_L = ½ LI², which is stored in the magnetic field established by the current.

3. Self inductance of a infinitely long straight air-cored solenoid is given by

L = μ₀n²Aℓ where ‘ℓ’ is the length of the solenoid, ‘A’ is its cross section area, ‘n’ is the number of turns per metre and μ₀ is the permeability of air (or free space).

[By the term infinitely long solenoid, we mean a solenoid with radius negligibly small compared to the length]

Note that the self inductance is proportional to the square of the number of turns.

4. Mutual inductance (coefficient of mutial induction) is defined in terms of magnetic flux linkage as

M = Ф/I where Ф is the magnetic flux (in weber) linked with the secondary circuit when a current I ampere flows in the primary circuit.

M is defined in terms of the induced voltage ‘ε’ in the secondary circuit as

M = ε /(dI/dt) where dI/dt is the rate of change of current in the primary circuit.

The induced voltage opposes the change in the current in the primary circuit in accordance with Lenz’s law and hence the above equation has to be strictly

M = – ε /(dI/dt). But remember that mutual inductance is a positive quantity.

It will be better to remember the above equation as

ε = –M(dI/dt).

5. Mutual inductance between an infinitely long straight solenoid and a short secondary coil wound round it (outside) at the middle is given by

M = μ₀nNA where 'n' and A are respectively number of turns per metre and the cross section area of the solenoid (primary) and N is the total number of turns in the secondary coil.

[If the secondary coil is inside the solenoid, the cross section area of the secondary coil is to be used in place of A].

6. Mutual inductance M is related to the primary and secondary self inductances L₁ and L₂ as

M = K√(L₁L₂) where K is the coupling coefficient which can have a maximum value of one. This happens when the entire magnetic flux produced by the primary is linked with the secondary.

Note that self inductance and mutual inductance are directly proportional to the permeability of the core material. If a coil is wound on a core of relative permeability μ_r, its inductance will be μ_r times the value with air core. Further, the self inductance of a given coil (with a given core) is a constant where as the mutual inductance between two given coils depends on the relative disposition of the coils.

7. Growth of current in a series LR circuit connected in series with a direct voltage is exponential and is given by

I = I₀(1– e^–Rt/L) where e is the base of natural logarithms and I₀ is the final current (at infinite time) given by I₀ = V/R, where V is the direct supply voltage.

8. Decay of current in the LR circuit is exponential and is given by

I = I₀e^–Rt/L

The growth and decay of current are shown in the adjoining figure.

9. Time constant of LR circuit = L/R

[You must remember the time constant of a CR circuit also, which is CR. The growth of charge Q on a capacitor connected in series with a direct voltage source of emf V volt is given by

Q= Q₀(1– e^–t/RC) where Q₀ is the final (maximum) charge equal to CV which is attained at infinite time.

When the capacitor having charge Q₀ is allowed to discharge through a resistance R, the decay of charge on the capacitor is given by

Q = Q₀e^–t/RC

The growth and decay of charge are shown in the adjoining figure.

10. Frequency of oscillations generated when a charged capacitor of capacitance C is discharged through an inductor of inductance L is given by

f = 1/2π√(LC)

The above equation can be written in terms of the angular frequency as

ω = 1/√(LC)

[The above equation is obtained by solving Kirchoff’s voltage equation for the LC circuit:

L(dI/dt) + Q/C =0, which can be rewritten as

L(d²Q/dt² ) + Q/C =0.

Therefore, d²Q/dt² = – Q/LC, which is similar to the equation of the simple harmonic motion of a mass ‘m’ attached to a spring of force constant ‘k’, written as

m (d²x/dt²) = – kx

or, d²x/dt² = –ω²x, where ω = √(k/m) is the angular frequency of mechanical oscillations of the spring-mass system.

The angular frequency of oscillations of the LC circuit (angular frequency of variation of charge on the capacitor) is similarly given by

ω = 1/√(LC)]

In the next post, questions (involving inductance) of the type you can expect in AP Physics C Exam will be discussed.

Answer to Free-Response Practice Question on Electromagnetic Induction

While answering free response questions never beat about the bush. Your effort should be to present things effectively in minimum words. At the same time, do not miss the required points. For instance, if you are asked to derive an expression, you should not write the final expression from your memory. You have to show the examiner, who evaluates your answers, that you know how to derive the expression, starting from basic things.

In the post dated 14^th December 2007, the following free response question on electromagnetic induction was given to you for practice:

The adjoining figure shows two smooth, straight, conducting, horizontal, parallel rails KL and MN on which a straight conducting rod PQ of mass ‘m’ can slide without friction, in a direction parallel to the rails. The rails and the rod have negligible resistance; but, there is a resistance R cunnected between the rails at one end. A uniform magnetic field of flux density B tesla is applied vertically downwards, throughout the region. The separation between the rails is L₁ and the length of the rod PQ is L₂.

(a) The rod PQ is pulled along the rails by a constant horizontal force F applied parallel to the rails. Derive an expression for the terminal velocity (v_terminal) of the rod.

(b) If the rod starts from rest with an initial acceleration ‘a₁’, find its acceleration when the velocity of the rod is v_terminal/4.

(c) What is the power dissipated in the resistance R in terms of the magnetic field and other known parameters, when the terminal velocity is attained?

(d) If the rod PQ is stationary and the magnetic field B is decreasing, will the direction of the induced current in the loop KPQM be clockwise or anticlockwise, when viewed along the direction of the magnetic field? Give reasons for your answer.

As promised, I give below the answer:

(a) The motional emf induced in the portion of the rod between the rails = BL₁v where ‘v’ is the velocity of the rod. (We have to consider the portion L₁ since the induced current will flow in this portion only).

Magnetic force on the rod = IL₁B where I is the induced current in the rod given by

I = BL₁v/R.

When terminal velocity v_terminal is attained, the applied force F is equal in magnitude (and opposite in direction) to magnetic force:

F = B²L₁² v_terminal/R, from which

v_terminal = FR/ B²L₁²

(b) Initially since v =0, there is no magnetic force and therefore, F = ma₁.

When the velocity is v_terminal/4, the magnetic force = B²L₁² v_terminal/4R

The net force = F – B²L₁² v_terminal/4R = ma.

Since F = B²L₁² v_terminal/R = ma₁ , the above equation becomes

ma₁ – ma₁/4 = ma from which a = 3a₁/4.

(c) The power dissipated (P) when the rod moves with terminal velocity is

P = I²R = (BL₁v_terminal /R)²×R = B²L₁² (v_terminal) ²/R.

(d) By Lenz’s law, the induced current in the loop should oppose the change (decrement here) in the magnetic flux. Therefore, the induced current in this case has to produce a magnetic field inthe same direction as the external field B. This happens if the current flows in the clockwise direction.