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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.

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 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.

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