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## Saturday, January 23, 2010

### AP Physics B- Additional Practice Questions (MCQ) on Geometric Optics

Some multiple choice practice questions on geometric optics were discussed in the posts dated 1st January 2008 and 3rd January 2008. You can access all posts related to geometric optics by clicking on the label ‘geometric optics’ below this post.

Today we will discuss a few more multiple choice practice questions in this section:

(1) The sun subtends an angle θ radian on the surface of the earth. If a convex lens of focal length F is used to obtain a real image of the sun on a screen, the diameter of the image will be

(a)

(b) Fθ/2

(c) F/θ

(d) 2

(e) impossible to be calculated with the given data.

Data is certainly sufficient for calculation. A very simple figure showing the formation of the image of the sun is shown. Since the sun is far away from the lens, its real image is formed at the focus of the lens. The angle subtended by this image at the optic centre of the lens is θ radian so that the diameter d of the image is .

[The sun subtends an angle of nearly 0.5º on the surface of the earth. If you use a convex lens of focal length 1 m, the diameter of the image will be = 1×(0.5×π/180), remembering that 180º = π radian. The diameter of the image will be nearly 0.0087 m = 0.87 cm].

(2) A slide projector gives a linear magnification of 20. If it projects a 4 cm×3 cm slide, the area of the image will be

(a) 0.24 m2

(b) 0.48 m2

(c) 4.8 m2

(d) 24 m2

(e) 48 m2

The length as well as the breadth will be magnified 20 times. In other words, the areal magnification is 20×20 = 400. Since the area of the slide is 12 cm2, the area of the image will be 12×400 = 4800 cm2 = 0.48 m2.

(3) Crown glass has refractive index 1.5. A cube of side 12 cm made of crown glass has a small air bubble inside and it appears to be at a distance of 6 cm when viewed normally through one face of the cube. If the air bubble is viewed normally through the opposite face, it will appear to be at a distance of

(a) 6 cm

(b) 5 cm

(c) 4 cm

(d) 3 cm

(e) 2 cm

In the case of normal refraction, the real distance dreal is related to the apparent distance dapp by the equation

n = dreal /dapp where n is the refractive index of the medium in which the object is placed. The refractive index n is with respect to the medium from which the object is viewed.

Therefore we have

1.5 = dreal /6 from which dreal = 9 cm.

Since the real distance of the air bubble from one face is 9 cm, the real distance from the opposite face is (12–9) cm = 3 cm.

Therefore the apparent distance x from the opposite face is given by

1.5 = 3/x from which x = 2 cm [Option (e)].

(4) A thin plano-convex lens has focal length 40 cm. The radius of curvature of its curved surface is 20 cm. If its curved surface is silvered, it can function as a concave mirror of focal length very nearly

(a) 6.67 cm

(b) 8.48 cm

(c) 10 cm

(d) 20 cm

(e) 32.8 cm

When light is incident on the plane face, it is refracted by the plano-convex lens. Then it is reflected by the concave mirror formed by the silvered curved face. While returning, once again it is refracted by the plano-convex lens. In effect the system behaves as the combination of two plano-convex lenses (each of focal length 40 cm) and a concave mirror (of focal length 10 cm since the radius of curvature is 20 cm), obeying the power relation,

1/F = 1/f1 + 1/f2 + 1/f3 where F is the combined focal length and f1, f2 and f3 are the individual focal lengths.

In the present case, f1 = f2 = 40 cm and f3 = 10 cm.

Therefore, 1/F = 1/40 + 1/40 + 1/10 = 1/20 + 1/10 = 3/20

This gives F = 6.67 cm, nearly.

[If the above lens is silvered on its plane face instead of the curved face, the focal length F of the equivalent concave mirror will be given by

1/F = 1/40 + 1/40 + 1/ since the focal length of a plane mirror is infinity.

Therefore, F = 20 cm].

## Saturday, January 16, 2010

### Answer to Free Response Practice Question on Geometric Optics for AP Physics B

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 axis of 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.

(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 appear to 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.

Now, find some multiple choice questions here.

## Thursday, January 14, 2010

### Geometric Optics- A Free Response Practice Question for AP Physics B

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 axis of 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.

(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.

## Friday, January 1, 2010

### Answer to Free Response Practice Question on Electromagnetism for AP Physics C

“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 and breadth b having negligible 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 sin ωt (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/r where ‘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 = 0I/r)ℓdr

The magnetic flux Ф through the entire wire loop is obtained by integrating the above expression between limits r = b and r = 2b.

Therefore, Ф =0Iℓ/2π) b2b(dr/r) = 0Iℓ/2π)[ln(2b) ln(b)]

Or, Ф =0Iℓ/2π) ln(2) = μ0Iℓ ln(2)/

(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×μ0Iℓ ln(2)/2π∆t = μ0Iℓ ln(2)/π∆t

(c) The magnetic flux linked with the wire loop when a current i = Im sin ωt flows 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 sin ωt) ln(2)/

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 = [μ0Imω ln(2) cos ωt]/

The maximum value of induced emf is μ0Imω ln(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 = [Cμ0Imω ln(2) cos ωt]/

The induced current iind in the wire loop is given by

iind = dQ/dt = [Cμ0Imω2 ln(2)( sin ωt)]/

The maximum value of induced current is Cμ0Imω2 ln(2) /2π.