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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Tuesday, November 12, 2013

### AP Physics B - Multiple Choice Practice Questions on Physical Optics

“One of the deep secrets of life is that all that is really worth the doing is what we do for others.”
– Lewis Carroll

Today we shall discuss a few typical multiple choice questions on physical optics:

Questions  (1) and (2) are based on the following statement:

A beam of monochromatic light having wave length λ, frequency f and intensity I in air enters a glass slab of refractive index 1.5. After traveling through the slab, the beam of light emerges through the opposite face of the slab and passes through air.

(1) Inside the slab the wave length and frequency of the light are respectively

(a) 1.5 λ and 1.5 f

(b) λ/1.5 and f

(c) 1.5 λ and f

(d) λ and f

(e)  λ and 1.5 f

The frequency of the light is unchanged where as the speed of the light is decreased within the slab. Note that the refractive index of a medium is the ratio of the speed of light in free space (or air at ordinary pressures) to the speed in the medium. Therefore, the speed of light (v) in the glass slab is given by

v = c/1.5 where ‘c’ is the speed of light in free space (or air).

The wave length λ, frequency f and speed v are related as

v = f λ

Or,       λ = v/f

Since the speed in glass is (1/1.5) times the speed in air, the wave length in glass is (1/1.5) times the wave length in air [Option (b)].

(2) The energy of the photons in glass is

(a) 1.5 times the energy in air

(b) 1/1.5 times the energy in air

(c) the same as the energy in air

(d) 3 times the energy in air

(e) 1/3 times the energy in air

The energy E of a photon is given by

E = hf where h is Planck’s constant and f is the frequency. Since the frequency of the light is the same in both air and glass, the energy of the photons in glass is the same as the energy in air [Option (c)].

(3) White light is passing through a transparent plastic slab. Inside the slab

(a) the green component  travels with maximum speed

(b) the green component  travels with minimum speed

(c) the violet component  travels with maximum speed

(d) the red component  travels with maximum speed

(e) all components travel with the same speed

The refractive index of any transparent medium is maximum for light of violet colour and minimum for light of red colour.

[This is why violet rays are deviated most and red rays are deviated least while traveling through a glass prism, producing dispersion of white light].

The speed of light (v) in the plastic slab is given by

v = c/n where ‘c’ is the speed of light in free space (or air) and n is the refractive inex of the slab.

Since the refractive index is the least for light of red colour, it follows that red component  travels with maximum speed [Option (d)].

(4) In an experiment with Young’s double slit, a student measures the intensity of the central maximum of the interference pattern as I. If one of the slits is covered, what will be the intensity at the position of the central maximum?

(a) I/2

(b) I/4

(c) I/(2)

(d) I

(e) 2I

When both slits are open, suppose the resultant amplitude (of light wave) at the position of the central maximum is a. If one of the slits is covered, the amplitude at the position of the central maximum becomes a/2.

Intensity is directly proportional to the square of the amplitude. Therefore we have

I α a2 and

I1 α (a/2)2

Therefore I1 = I/4, as given in option (b).

(5) Four laser sources produce light waves y1, y2, y3, and y4 given (with usual notations) by

y1 = a sin ωt

y2 = a sin 2ωt

y3 = 2a sin (ωt + φ)

y4 = 2a sin (3ωt + φ)

Superposition of which two waves can produce interference fringes?

(a) y1 and y2

(b) y2 and y3

(c) y3 and y4

(d) y1 and y4

(e) y1 and y3

Waves y1 and y3 have the same wave length and hence they can produce interference fringes.

[The angular frequencies of y1 and y3 are equal].

The correct option is (e).

You will find a few useful multiple choice questions in this section here.

## Friday, September 13, 2013

### AP Physics B & C - Electrostatics - Multiple Choice Practice Questions

“Iron rusts from disuse, stagnant water loses its purity and in cold weather becomes frozen; so does inaction sap the vigors of the mind.”
– Leonardo da Vinci

Questions on electrostatics were discussed on many occasions on this site. You can access them by clicking on the label ‘electrostatics’ below this post or by trying a search for ‘electrostatics’ using the search box provided on this page. Today we shall discuss a few more questions in this section.
(1) Right angled triangle ABC is located in a uniform electric field E. Sides AC and BC have lengths 0.5 m and 0.3 m respectively and BC is at right angles to the electric field lines (Fig.). If the electric potential difference between A and C is 80 V, what is the magnitude of the electric field E?
(a) 80 V/m
(b) 160 V/m
(c) 200 V/m
(d) 100 V/m
(e) 16 V/m
The electric potential at the point B is the same as that at the point C (since the straight line BC is at rjght angles to the direction of the uniform electric field E). Therefore the potential difference between points A and B is 80 V. The electric field E is directed along AB and hence the magnitude of E is (80 V)/(0.4 m) = 200 V/m.
[AB = √(AC2 – BC2) = √(0.52 – 0.32)  = 0.4 m]
(2) In the above question, what is the component of electric field along the direction AC?
(a) 80 V/m
(b) 160 V/m
(c) 200 V/m
(d) 100 V/m
(e) 40 V/m
The potential difference between the points A and C is 80 V and the distance between these points is 0.5 m. Therefore, the component of electric field along the direction AC is (80 V)/(0.5 m) = 160 V/m.

(3) A cube of side a has a charge Q at its centre (Fig.). What is the electric flux through one face of the cube?
(a) Q/ε0a
(b) Qa/ε0
(c) Q/6ε0
(d) Q/8ε0
(e)0/6
The electric flux over a closed surface, according to Gauss theorem, is Q/ε0 where Q is the net charge enclosed by the surface and ε0 is the permittivity of free space.
[You can understand the above even without knowing Gauss law. You know (from inverse square law) that the electric field at distance r from a point charge Q is Q/4πε0r2. The electric field at any point is the electric flux through unit area held with the plane of the area perpendicular to the electric field lines. Imagine a spherical surface of radius r such that the charge Q is at the centre. Since the area of the spherical surface is 4πr2 the electric field at any point on the surface must be Ф/4πr2 where Ф is the total electric flux produced by the charge Q. Therefore we have
Q/4πε0r2 = Ф/4πr2
This gives Ф = Q/ε0 as stated in Gauss law.]
Since the charge Q is at the centre of the cube, all the six faces of the cube receive equal electric flux so that the flux through one face of the cube is Ф/6 = Q/6ε0

(4) The effective capacitance between terminals A and B in the network shown in the adjoining figure is
(a) 16 μF
(b) 8 μF
(c) 6 μF
(d) 16/3 μF
(e) 8/3 μF
Since the capacitors C1, C2, C3 and C4 are of values satisfying the condition C1/C2 = C3/C4, the junction of C1 and C2 is at the same potential as the junction of C3 and C4 (as in the case of a balanced Wheatstone brige). Therefore the capacitors C5 and C6 connected across the diagonal of the bridge have no effect and can be ignored.
The network therefore simplifies to the series combination of C1 and C2 connected in parallel with the series combination of C3 and C4.
Series combined value of C1 and C2 = (3×6)/(3+6) = 2 μF
Series combined value of C3 and C4 = (1×2)/(1+2) = 2/3 μF
Therefore the effective capacitance between terminals A and B = 2 μF + (2/3) μF = 8/3 μF

The capacitors C1 and C4 in the above question are short circuited and the circuit then gets modified as shown in the figure. What is the effective capacitance between the terminals A and B in this situation?
(a) 11 μF
(b) 9 μF
(c) 8 μF
(d) 7.5 μF
(e) 6.5 μF
Have a careful look at the circuit. You will find that one plate of C2, C3 and C5 is connected to terminal A. The other plate of C2 as well as C3 is connected to terminal B while that of C5 is connected 6through C6 to terminal B. The series combination of C5 and C6 gives an effective capacitance of 1μF. Thus we have three capacitances 1μF, 1μF and 6μF connected in parallel across the terminals A and B.

Therefore the effective capacitance between terminals A and B in this case is 1μF + 1μF + 6μF = 8μF.

## Friday, August 9, 2013

### AP Physics B - Multiple Choice Practice Questions on Geometric Optics

– Mark Twain

Questions on geometric optics were discussed on many occasions on this site. Try a search for ‘geometric optics’ using the search box provided on the side bar this page or click on the label ‘geometric optics’ below this post, to access all posts related to geometric optics.
Today we shall discuss a few more multiple choice practice questions on geometric optics:
(1) Focal power or optical power (or simply, power) of a lens is the reciprocal of its focal length. The focal length has to be expressed in metre for obtaining the power in its popular unit, dioptre.
Two thin converging lenses of focal lengths 40 cm and 60 cm are kept in contact so as to have a common principal axis. The power of this combination is approximately
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
If the focal length of the combination of the lenses is F metre, the power of the combination is 1/F.
When two thin lenses of focal lengths f1 and f2 are kept in contact so as to have a common principal axis, the focal length F of the combination is given by the reciprocal relation,
1/F = 1/f1 + 1/f2
[In fact, this is a relation connecting the powers of the individual lenses to the power of the combination]
Therefore we have 1/F = 1/0.4 + 1/0.6 = (0.6 + 0.4)/(0.4×0.6) = 1/0.24 ≈ 4

(2) A bright object O is placed on the principal axis of a thin converging lens. The lens  produces a real magnified image of O at the position I (Fig.) at a distance of 44 cm from the lens. When a convex mirror of focal length 12 cm is interposed between the image and the lens as shown in the figure, a real image of the same size as the object is obtained side by side with the object. What is the distance between the lens and the mirror?
(a) 12 cm
(b) 16 cm
(c) 20 cm
(d) 22 cm
(e) 24 cm
Since the final image is obtained side by side with the object, the rays of light must fall normally on the mirror. This means that the initial image I must be formed at the centre of curvature of the mirror. The radius of curvature of the mirror is twice its focal length and is therefore equal to 24 cm. The initial image I must therefore be at a distance of 24 cm from the mirror. Since the distance between the lens and the initial image I is 44 cm, the distance between the lens and the mirror must be (44 – 24) cm = 20 cm.
(3) An object is placed in front of a convex mirror of focal length f . The distance of the object is greater than f  but less than 2f. The image of the object is
(a) in front of the mirror at distance less than f
(b) in front of the mirror at distance greater than f  but less than 2f
(c) in front of the mirror at distance 2f
(d) behind the mirror at distance less than f
(e) behind the mirror at distance greater than f  but less than 2f
You may try drawing a ray diagram to locate the image, as shown in the adjoining figure. A ray of light proceeding towards the centre of curvature C of the mirror gets reflected from the mirror and retraces its path. Another ray proceeding parallel to the principal axis gets reflected from the mirror and proceeds as shown, as though it diverges away from the focus F of the mirror. The image is virtual and is located behind the mirror at distance less than f.
The correct option is (d).
[Note that the image will be located behind the mirror at distance less than f for all object distances].
(4) A student tries to burn a sheet of paper by focusing sun light on it using a converging lens of focal length f. The radius of the sun is R and its distance from the earth is D. The diameter of the image of the sun obtained on the sheet of paper is
(a) 2Rf/D
(b) 2RD/f
(c) RD/f

(d) RD/2f
(e) Rf/2D
Since the sun is far away from the lens, a real diminished image of the sun is formed at the focus of the lens.
[The student has to keep the sheet of paper at the focal plane of the lens for burning it].
The formation of the image is shown in the adjoining figure in which the radius of the image is shown as r.
From the similar triangles, we have
R/r = D/f
Or, r = Rf/D
The diameter of the image = 2r = 2Rf/D