A free response practice question involving geometric optics was given to you in the post dated 14th January 2010. As promised, I give below a model answer along with the question:
The figure shows a thin converging lens of focal length 12 cm. A small object O (indicated by a vertical arrow) is placed on the principal axisof this lens at a distance of 10 cm from the optic centre of the lens.
(a) Making use of at least two rays proceeding from the object, draw a ray diagram showing the image formed by the lens.
(b) Comment on the nature of the image: whether it is real or virtual; magnified or diminished; erect or inverted.
Justify your comment.
(c) Another thin converging lens of focal lens 20 cm is kept in contact with the above lens so that they have a common principal axis. The object O is kept at 10 cm itself from the centre of the combination of these lenses. Calculate the distance of the image (formed by this combination) from the centre of the lens system, making use of the law of distances.
(d) Is the image formed in this case real or virtual? Justify your answer.
Try to answer the above question. You can take about 11 minutes for answering it and can score up to 10 points for the right answer. I’ll be back shortly with a model answer for you.
(a) The ray diagram is shown in the following figure:
To draw the ray diagram two rays are considered: One ray proceeding parallel to the principal axis gets refracted at the lens and passes through the principal focus F. Another ray proceeding through the optic centre of the lens is undeviated.
(b) The image is virtual since the rays do not rally converge at the image, but only appearto diverge from it.
The image is magnified and erect as is evident from the ray diagram.
(c) When the two lenses are in contact, the focal length F of the combination is given by
1/F = 1/f1 + 1/f2 = 1/12 + 1/20
Therefore, F = (12×20)/(12 + 20) = 7.5 cm
The distance (si) of the image formed by the combination of the lenses is given by
1/F= 1/si– 1/so where so is the object distance (10 cm)
Substituting for F and so, we have
1/7.5 = 1/si– 1/(–10)
[The object distance is negative in accordance with the Cartesian sign convention]
Therefore, 1/si = 1/7.5 – 1/10 from which si = 30 cm.
(d) The image is real since the image distance is positive.
The essential points to be remembered in geometric optics were discussed in the post dated 29th December 2007. You can access all posts related to geometric optics by clicking on the label ‘geometric optics’ below this post.
Today I give you a free response practice question in this section:
The figure shows a thin converging lens of focal length 12 cm. A small object O (indicated by a vertical arrow) is placed on the principal axisof this lens at a distance of 10 cm from the optic centre of the lens.
(a) Making use of at least two rays proceeding from the object, draw a ray diagram showing the image formed by the lens.
(b) Comment on the nature of the image: whether it is real or virtual; magnified or diminished; erect or inverted.
Justify your comment.
(c) Another thin converging lens of focal lens 20 cm is kept in contact with the above lens so that they have a common principal axis. The object O is kept at 10 cm itself from the centre of the combination of these lenses. Calculate the distance of the image (formed by this combination) from the centre of the lens system, making use of the law of distances.
(d) Is the image formed in this case real or virtual? Justify your answer.
Try to answer the above question. You can take about 11 minutes for answering it and can score up to 10 points for the right answer. I’ll be back shortly with a model answer for you.
“The object of a New Year is not that we should have a new year. It is that we should have a new soul.”
– G. K. Chesterton
Happy New Year…
A free response practice question involving electromagnetism was given to you in the post dated 30th December 2009. As promised, I give below a model answer along with the question:
A rectangular wire loop of length ℓ and breadth b havingnegligible resistance is arranged in the plane of an infinitely long straight vertical wire as shown, with the longer sides parallel to the wire. The wire loop contains a capacitor of capacitance C. A steady current ‘I’ flows upwards through the straight wire. Now answer the following questions:
(a) Calculate the magnetic flux through the wire loop.
(b) The loop is rotated through 180º about a central axis (fig.) which is parallel to the straight wire. If the time taken for this rotation is ∆t, determine the magnitude of the emf induced in the loop.
(c) The loop is kept stationary and instead of the steady current I, a current ‘i’varying with time t as i = Im sinωt (where Im and ω are constants) is passed through the straight wire. Calculate the maximum value of the emf induced in the wire loop.
(d) Determine the maximum current induced in the wire loop under the conditions mentioned in part (c) above.
(a) The magnetic field B (produced by the straight current carrying wire) at the loop varies with distance as
B = μ0I/2πrwhere ‘r’ is the distance of the point from the straight wire.
The magnetic field at the wire loop is directed normally in to the plane of the loop. The magnetic flux through the entire wire loop can be found considering strips parallel to the straight wire. One such strip of width dr at distance r is shown in the figure. This strip has area dA = ℓdr and hence the magnetic flux through this strip is given by
dФ =BdA = (μ0I/2πr)ℓdr
The magnetic flux Ф through the entire wire loop is obtained by integrating the above expression between limits r = band r = 2b.
(b) When the wire loop is rotated through 180º, the magnetic flux through the loop changes from Ф to –Ф so that the change of flux is 2Ф. The magnitude of the emf induced in the wire loop is equal to the time rate of change of the magnetic flux through the loop and is equal to 2Ф/∆t = 2×μ0Iℓln(2)/2π∆t = μ0Iℓln(2)/π∆t
(c) The magnetic flux linked with the wire loop when a current i = Im sinωtflows through the straight wire is obtained by replacing I by (Im sin ωt) in the expression for Ф obtained in part (b) above.
Thus Ф = μ0(Im sinωt)ℓln(2)/2π
The magnitude of the emf induced in the wire loop is equal to dФ/dt.
[The induced emf is – dФ/dt, the negative sign appearing because of Lenz’s law. We ignore the negative sign since we are interested in the magnitude of the emf].
Therefore, induced emf, V = [μ0Imωℓln(2)cosωt]/2π
The maximum value of induced emf is μ0Imωℓln(2)/2π, appropriate to the maximum value of 1 for cos ωt.
(d) The charge Q on the capacitor because of the induced emf is given by
Q = CV = [Cμ0Imωℓln(2)cosωt]/2π
The induced current iindin the wire loop is given by
iind = dQ/dt = [Cμ0Imω2ℓln(2)(–sinωt)]/2π
The maximum value of induced current is Cμ0Imω2ℓln(2)/2π.
You can access all posts involving magnetic fields and electromagnetic induction on this blog by trying a search using the search box at the top of this page or by clicking on the labels ‘magnetic field’ and ‘electromagnetic induction’ below this post. Today I give you a free response practice question on electromagnetism. This question is for AP Physics C aspirants:
A rectangular wire loop of length ℓ and breadth b havingnegligible resistance is arranged in the plane of an infinitely long straight vertical wire as shown, with the longer sides parallel to the wire. The wire loop contains a capacitor of capacitance C. A steady current ‘I’ flows upwards through the straight wire. Now answer the following questions:
(a) Calculate the magnetic flux through the wire loop.
(b) The loop is rotated through 180º about a central axis (fig) which is parallel to the straight wire. If the time taken for this rotation is ∆t, determine the magnitude of the emf induced in the loop.
(c) The loop is kept stationary and instead of the steady current I, a current ‘i’varying with time t as i = Im sinωt (where Im and ω are constants) is passed through the straight wire. Calculate the maximum value of the emf induced in the wire loop.
(d) Determine the maximum current induced in the wire loop under the conditions mentioned in part (c) above.
Try to answer this question. You can take 15 minutes for answering it and can score up to 15 points for the right answer. I’ll be back soon with a model answer for your benefit.
Equations to be remembered in the section ‘oscillations’ were discussed in the post dated 17thApril 2008.Some multiple choice practice questions in this section were discussed in the post dated 22nd April 2008, followed by a free response practice question in the post dated 2ndMay 2008. You can access all these posts by clicking on the label ‘oscillation’ below this post. Today we will discuss a few multiple choice practice questions involving simple pendulum:
(1) A simple pendulum arranged inside a train has period of oscillation T when the train is at rest. When the train moves along straight horizontal rails with uniform acceleration of x ms–2, the period of the pendulum is (assuming g = 10 ms–2)
(a) T ×g1/2/(g+x)1/2
(b) T ×g1/4/(g+x)1/4
(c) T ×g1/4/(g2+x2)1/4
(d) T ×g1/2/(g2+x2)1/2
(e)T ×g1/2/(g2+x2)1/4
When the train is at rest the period of oscillation T of the pendulum is given by
T = 2π (ℓ/g)1/2 ………….(i)
where ℓ is the length of the pendulum and g is the acceleration due to gravity.
When the train moves forward with acceleration x ms–2, an inertial backward force acts on the bob of the pendulum and supplies a backward acceleration of x ms–2 (fig.). The resultant acceleration of the bob is (g2+x2)1/2. The period T1of the pendulum is now given by
T1 =2π[ℓ/(g2+x2)1/2]1/2……..(ii)
Dividing eqn (ii) by eqn (i) we have
T1/T = g1/2/(g2+x2)1/4 from which T1 = T ×g1/2/(g2+x2)1/4 The correct option is (e)
[Suppose the train is moving with uniform velocity of x ms–1. What will be the period? No doubt, T itself since you cannot distinguish between state of rest and uniform motion].
(2) Two simple pendulums A and B have periods 2.1 s and 2 s respectively. They start oscillating at the same time in phase. They will be in phase instantly at the end of
(a) 42 s
(b) 40 s
(c) 22 s
(d) 21 s
(e) 20 s
Suppose the pendulums are in phase instantly at the end of n oscillations of pendulum A. Pendulum B should then execute n+1 oscillations. Therefore we have
n×2.1 = (n+1)×2
Therefore 0.1n = 2 so that n = 20
The time elapsed is therefore 20×2.1 = 42 s [Option (a)].
[The difference between the periods of the two pendulums is 0.1 s. So you require 20 oscillations of pendulum A to obtain a time difference equal to one period (2 s) of pendulum B so that the two pendulums will be instantly in phase].
(3) The bob of a simple pendulum is a hollow metal sphere filled with water. There is a small hole at the bottom of this hollow sphere and water drains out through the hole as the pendulum oscillates. After completely filling the bob with water, if this pendulum is made to oscillate for a long time, its period of oscillation will
(a) increase first and will reach a final constant value
(b) decrease first, reach a minimum value, then increase and will finally settle at the initial value
(c) increase first, reach a maximum value, then decrease and will finally settle at the initial value
(d) decrease first and will reach a final constant value
(e) remain unchanged
The correct option is (c). Initially the centre of gravity of the spherical bob is at its centre since it is completely filled with water. When the water flows out, the centre of gravity of the bob moves gradually downwards, reaches a minimum level and then moves up. When the water is fully drained out, the centre of gravity of the bob once again reaches the centre of the bob and remains there.
The length of the pendulum is the distance between the centre of gravity of the bob and the point of suspension. Therefore, the length of the pendulum gets increased initially, becomes a maximum, then gets decreased and finally settles at the initial value. Therefore, the period of the pendulum will increase first, reach a maximum value, then decrease and will finally settle at the initial value.
(4) A simple pendulum is taken to a location where the acceleration due to gravity is decreased by 0.1 %. If the period of oscillation is to be unaltered
(a) the mass of the bob is to be increased by 0.1 %
(b) the mass of the bob is to be decreased by 0.1 %
(c) the length of the pendulum is to be decreased by 0.316 %
(d) the length of the pendulum is to be decreased by 0.2 %
(e) the length of the pendulum is to be decreased by 0.1 %
Since the period of oscillation T is given by T = 2π√(ℓ/g), the ratio ℓ/g should be unaltered. Therefore, the length ℓ should be decreased by 0.1 %
The following questions are for AP Physics C aspirants only:
(5) The period of a simple pendulum is decreased by 0.02 s when the length of the pendulum is decreased by 1 cm. The original length of the pendulum is nearly (g = 10 ms–2)
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(a) 0.25 m
(b) 0.5 m
(c) 1 m
(d) 1.01 m
(e) 2 m
The period of oscillation T is given by T = 2π√(ℓ/g).
Therefore, ∆T/T = ½ ∆ℓ/ℓ – ½ ∆g/g
[Here ∆T,∆ℓ, and∆g represent the increments in the period, length and acceleration due to gravity respectively. We have written the above equation by taking the logarithm of the expression for period and then differentiating it]
The increment in T is – 0.02 s and the increment in ℓ is – 0.01 m (negative signs are because T and ℓ are decreased). There is no change in g.
Therefore we have – 0.02/T = – ½ × 0.01/ℓ
This gives T = 4ℓ.
But T = 2π√(ℓ/g) so that 2π√(ℓ/10) = 4ℓ
Or, π2/10 = 4ℓ
Since π2 is nearly equal to 10, we obtain ℓ = 0.25 m, nearly.
