AP Physics B - Multiple Choice Practice Questions on Geometric Optics

“A person who won't read has no advantage over one who can't read.”

– 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  

AP Physics C – Magnetic Fields – Answer to Free Response Practice Question on Magnetic Force on Moving Charges

A free response practice question on magnetic force on moving charges was posted on 8^th July 2013. As promised I give below a model answer for your benefit, along with the question:

The adjoining figure represents a cathode ray tube in a cathode ray oscilloscope. The voltage applied (with respect to the cathode) to the final accelerating anode is V and the distance between the anode and the fluorescent screen of the cathode ray tube is L. The mass of the electron is m and the magnitude of its charge is e.  The cathode ray tube is so oriented that in the undeflected condition the electron beam is horizontal and is along the magnetic meridian, with the electrons traveling from south to north. The horizontal component of the earth’s magnetic field at the place is B_h. Now, answer the following questions:

(a) Calculate the velocity with which the electrons strike the fluorescent screen, assuming that the space between the accelerating anode and the fluorescent screen is devoid of electric fields and the electrons start from the cathode with negligible speed.

(b) The cathode ray tube is rotated through 90º about a verti8cal axis so that the electrons travel from east to west. In what direction will the spot of light (produced by the impact of the electron beam) on the fluorescent screen be deflected? Select the correct option from the following, by putting a tick mark (√) against one of the following:

Upwards………

Downwards………

Leftwards………

Rightwards……..

Justify your answer.

(c) If the cathode ray tube is rotated through 90º in the opposite direction, will the direction of deflection of the spot of light on the screen change? Justify your answer.

(d) Derive an expression for the specific charge (charge to mass ratio) of the electron in terms of the given data and the shift s of the electron spot on the screen, on rotating the cathode ray tube through 90º. Assume that the shift s is small compared to the distance between the accelerating anode and the fluorescent screen.

Answer:

(a) The electrons acquire kinetic energy under the action of the accelerating potential V applied on the accelerating anode. If the final velocity acquired by the electrons is v, we have

             ½ mv² = eV

Therefore v = √(2eV/m)

This is the velocity with which the electrons strike the fluorescent screen since their velocity after leaving the aperture of the accelerating anode remains unchanged in the field free space between the anode and the screen.

(b) Upwards.

Earth’s magnetic field is directed from south to north. Electrons traveling from east to west make a conventional current flowing from west to east (since the electrons are negatively charged).

[The current produced by electrons traveling from east to west has the same direction as the current produced by positive charges traveling from west to east]

Therefore, in order to apply Fleming’s left hand rule for fining the direction of the magnetic force on the electron beam, we have to hold the forefinger along the south to north direction and the middle finger along the west to east direction. The thumb will then point upwards.

(c) If the cathode ray tube is rotated through 90º in the opposite direction, the electron beam will be deflected downwards since the electrons  travel from west to east.

(d) When the electrons travel parallel to the horizontal component (B_h) of earth’s magnetic field, there is no magnetic force on them. This follows from the expression for magnetic force F:

             F =evB_h sinθ where θ is the angle between the magnetic field and the direction of the conventional current. Since θ = 180º, F = 0.

When the cathode ray tube is rotated through 90º, the electrons travel at right angles to the horizontal component (B_h)  of earth’s magnetic field. In this case the magnetic force is maximum (F_max) and is given by

             F_max = evB_h. 

Since the deflection (s) produced is small (Fig.), we may assume that this force acts vertically throughout the path of the electrons.

The above vertical force produces a vertical acceleration and the vertical displacement s suffered by the electrons is given by the equation of uniformly accelerated motion,

             s = ut + ½ at² ……….. (i)

where u is the initial vertical velocity (at the instant of leaving the accelerating anode), which is zero, t is the time taken by the electron to travel from the accelerating anode to the fluorescent screen and a is the vertical acceleration which is given by

             a = F_max/m = evB_h/m  

The time taken by the electron to travel the distance L between the anode and screen is given by

             t = L/v

Substituting for u, a and t in Eq. (i) we have

             s = 0 + ½ (evB_h/m) (L/v )² = ½ (e/m) (B_hL²/v)

The speed v of the electrons, as shown in part (a) is √(2eV/m)

Substituting for v we have

             s = ½ (e/m) (B_hL²/√(2eV/m)

This gives e/m = 8s²V/ B_h²L⁴

AP Physics C – Magnetic Fields – Free Response Practice Question on Magnetic Force on Moving Charges

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

– Lewis Carroll

 

The cathode ray oscilloscope is one of the most versatile instruments which is used as a test instrument in a variety of forms in almost all fields of scientific activity. Most of you may be familiar with this impressive instrument which is essentially a highly evacuated glass bulb containing an electron gun and a fluorescent screen, with provisions for deflecting the electron beam, usually by applying voltages across deflecting plates. I give below a free response question related to the cathode ray tube:

The adjoining figure represents a cathode ray tube in a cathode ray oscilloscope. The voltage applied (with respect to the cathode) to the final accelerating anode is V and the distance between the anode and the fluorescent screen of the cathode ray tube is L. The mass of the electron is m and the magnitude of its charge is e.  The cathode ray tube is so oriented that in the undeflected condition the electron beam is horizontal and is along the magnetic meridian, with the electrons traveling from south to north. The horizontal component of the earth’s magnetic field at the place is B_h. Now, answer the following questions:

(a) Calculate the velocity with which the electrons strike the fluorescent screen, assuming that the space between the accelerating anode and the fluorescent screen is devoid of electric fields and the electrons start from the cathode with negligible speed.

(b) The cathode ray tube is rotated through 90º about a verti8cal axis so that the electrons travel from east to west. In what direction will the spot of light (produced by the impact of the electron beam) on the fluorescent screen be deflected? Select the correct option from the following, by putting a tick mark (√) against one of the following:

Upwards………

Downwards………

Leftwards………

Rightwards……..

Justify your answer.

(c) If the cathode ray tube is rotated through 90º in the opposite direction, will the direction of deflection of the spot of light on the screen change? Justify your answer.

(d) Derive an expression for the specific charge (charge to mass ratio) of the electron in terms of the given data and the shift s of the electron spot on the screen, on rotating the cathode ray tube through 90º. Assume that the shift s is small compared to the distance between the accelerating anode and the fluorescent screen.

Try to answer this question. You are required to answer it within 15 minutes and it carries 15 points.

I’ll be back soon with a model answer for your benefit.

AP Physics B – Simple and Interesting Fluid Mechanics

“There is no democracy in physics. We can't say that some second-rate guy has as much right to an opinion as Fermi.”

–Luis Walter Alvarez

The first free response question in the AP Physics B 2013 Examination was from fluid mechanics which provides ample scope for question setters to ask interesting questions. Today we shall discuss some interesting and useful points in this section.

An important point you need to note is that an object denser than a liquid, can displace liquid of  weight equal to that of the object, if the object forms part of a floating  body.  The same object submerged independently in the liquid will displace liquid of weight less than that of the object.

(1) To make the above point more clear, let us consider a large tank containing water of density 1000 kg/m³. A wooden block A of specific gravity 0.5 and volume 2 m³ is floating in the water (Fig. 1). An object B of specific gravity 5 and volume 0.1 m³ is placed on the wooden block. Will the wooden block still continue to float?

The wooden block A has mass 1000 kg as is clear from the relation,

             mass = volume×density

[Since the specific gravity of the wooden block is 0.5, its density is 0.5×1000 = 500 kg/m³. Therefore, the mass of the wooden block = 2×500 = 1000 kg.

Remember that the specific gravity or relative density of a body is the ratio of the density of the body to the density of water. Since water has density 1000 kg/m³ in the international system (SI) of units, the density of a body has numerical value 1000 times its specific gravity].

Mass of the object B placed on the wooden block is 0.1×5000 = 500 kg. Therefore the total mass of the wooden block and the object placed on it is (1000+500) kg = 1500 kg.

Since the volume of the wooden block is 2 m³ it can displace 2 m³ of water before getting submerged. The wooden block and the object placed on it will continue to float since the total volume of water to be displaced needs to be 1.5 m³ only.

[Mass of 1.5 m³ of water is 1500 kg].                                                                                    

In the above explanation we have used the law of floatation, which states that the weight of a floating body is equal to the weight of the displaced liquid.

(2) Now, consider a slightly different situation which is shown in the adjoining figure (Fig. 2). The same wooden block A and the denser object B as we considered above are shown here also; but the object is suspended from the wooden block using a rope of negligible mass. How much water will be displaced by the system (containing the wooden block and the object) in this case?

The object has specific gravity 5 and it is fully immersed in water. If it were not part of the floating system, it would have displaced water of volume equal to its own volume (which is 0.1 m³). Since the object is part of a floating system, it is capable of displacing water of mass equal to its own mass (which is 500 kg), by pulling the wooden block further down making use of the connecting rope. Thus the total mass of water displaced is 1500 kg and the total volume of water displaced is 1.5 m³, as in the case of the system shown in Fig. 1.

In Fig. 1 the water level in the tank is shown as L₁ where as in Fig. 2 the water level is shown as L₂. Obviously these two levels coincide.

(3) Now, suppose the rope connecting the object B to the wooden block A cut. The object B, being denser than water, will then sink and will rest at the bottom of the tank. How much water will be displaced by the wooden block and the object together in this case?

The wooden block is a floating body and hence it will displace water having mass equal to its own mass, which is 1000 kg. So the volume of water displaced by the wooden block is 1 m³. The object B will displace water of volume equal to its own volume, which is equal to 0.1 m³. Therefore, the total volume of water displaced by the wooden block and the object together in this case is 1.1 m³.

The water level in the tank will be lowered (compared to L₁) in this case.

(4) Let us once again consider the original system (Fig. 1) in which the denser object B (of specific gravity 5 and volume 0.1 m³) is placed on the wooden block. What must be the mass of the object B if the entire wooden block just begins to submerge?

Questions of the above type are often found in question papers of various examinations. When the wooden block is completely immersed in water the volume of water displaced is equal to the volume of the wooden block, which is 2 m³. The weight of the wooden block and the object together in this case is therefore equal to the weight of 2 m³ of water, which is 2000 kg. Since the mass of the wooden block is 1000 kg, the mass of the object must be 1000 kg.

We shall now consider another situation:

(5) Suppose the object placed on the wooden block (Fig. 1) has specific gravity 0.8 and the water level in the tank is noted. If the object is gently transferred to the water in the tank, will the water level in the tank change?

Since the specific gravity of the object is 0.8, it will float in water. Whether it is on the wooden block or independently outside, it will displace water of mass equal to its own mass. Therefore, there is no change in the water level in the tank.

(6) Now, suppose the object of specific gravity 0.8 (which we considered above) is tied to the wooden block using a rope of negligible mass so that it is fully immersed in the water in the tank, as shown in Fig. 3. Will the level of water in the tank change in this case?

Many among you might be confused about this situation. Understand that the wooden block and the object tied to it is still a floating system and the mass of water displaced must be equal to the mass of the floating system. Therefore there is no change in the water level in the tank.

We shall now consider the case of an ice block floating in water: 

(7) Consider a block of ice floating in a tank containing pure water (Fig. 4). If the ice melts without any change in the temperature, will the water level in the tank rise up, fall down or remain unchanged?

This is a very popular question and most of you know the answer: The water level will remain unchanged. But how do you justify your answer?

The block of ice is a floating body and the volume of water displaced by it has weight equal to its own weight. When the ice melts, the eaxtra water produced should fill exactly the volume originally occupied by the immersed portion of ice. Therefore, the water level in the tank is unchanged.

(8) Suppose a metallic bob is placed on the ice block so that it still floats in the water in the tank (Fig. 5). If the ice melts without any change in the temperature, will the water level in the tank change?

When the ice alone is there, the water level in the tank remains unchanged when the ice melts, as we have seen above. When the metallic bob also is present, it sinks when the ice melts and it displaces water of volume equal to its own volume. This volume is certainly less than the volume of water it displaces when it is part of the floating system formed along with the ice block. The water level in the tank is therefore lowered when the ice melts (without any change in the temperature).

(9) Now, suppose the metallic bob in the above question is replaced by a wooden bob (which is lighter than water). If the ice melts without any change in the temperature, will the water level in the tank change?

In this case there will be no change in the water level since the wooden bob floats in the water in the tank when the ice melts, displacing the same volume of water as before.