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## Tuesday, January 1, 2008

### AP Physics B- Multiple Choice Questions from Geometric Optics

The following multiple choice questions (MCQ) will help to boost your morale in facing the AP Physics B Exam:
(1) At what angle with respect to the horizontal, does a fish under water see the setting sun?
(a) tan–1(7/3) (b) tan–1(5/3) (c) tan–1(5/7)
(d) tan–1(5/3) (e) tan–1(7/4)

With reference to the figure, S represents the setting sun from which the ray SAF reaches the fish F after refraction at F. For the fish the ray appears to come from S1 which is the virtual image of the sun. Evidently angle FAN is the critical anglec) for the water-air interface. Since n = 1/sinθc, we have
sinθc = 1/n = 1/(4/3) =3/4
[Right angled triangle FAB is drawn side by side with the figure showing the refraction, to show the opposite side of length 3 units and adjacent side of length 4 units. The hypotenuse of this triangle is √7].
The rays coming from the sun make an angle S1AS = FAB with the horizontal. The angle required in the problem is angle FAB, which is tan–1(√7/3).
(2) In the adjoining figure, SS is a spherical surface separating two media of refractive indices n1 and n2 where n1 > n2. C is the centre of curvature of the spherical surface. An observer, keeping his eye beyond C in the medium of refractive index n2 views the refracted image of an object AB placed as shown in the medium of refractive index n1. The image will be (a) real, inverted and magnified
(b) real, upright and diminished
(c) virtual, upright and diminished
(d) virtual, upright and magnified

You can easily solve this problem by ray tracing as shown in the figure.
CN is normal to the spherical surface. A ray BC passing through the centre of curvature C does not suffer any deviation. Another ray BD is deviated at D obeying Snell’s law of refraction, n1 sin θ1 = n2 sin θ2, where θ1 is the angle of incidence BDN and θ2 is the angle of refraction EDC. In this question you are asked to find the nature of the image qualitatively and you need not worry about the actual values of θ1 and θ2. But you must be careful to draw the refracted ray DE so as to have θ2 larger than θ1 since n1 > n2.
The diverging ray DE produced backwards meets the ray BC at M and IM is the virtual, upright and diminished image of the object AB. So, option (c) is correct.
Suppose you were asked to calculate the distance of the image and its magnification. In that case, the refractive index values, the radius of curvature of the spherical surface and the distance of the object will be given in the problem. You will have to work out things quantitatively. If you still want to resort to the method of ray tracing, you should draw all distances to a suitable scale. The spherical surface SS is to be first drawn by fitting the radius to the chosen scale. The object AB is to be shown at the given distance (to the scale). After drawing the ray BD you will have to measure the angle of incidence θ1 ( angle BAN) and then calculate angle of refraction θ2 (angle EDC) using n1 sin θ1 = n2 sin θ2. The refracted ray DE is to be drawn so that angle EDC is equal to the calculated value of θ2. After completing the ray diagram, you will have to measure the distance of the image (PI). The size of the image (IM) need not be measured since you can calculate magnification (M) using the equation M = si/so.
If a question does not demand you to do things specifically by ray tracing, you can calculate all quantities using the equations we considered in the post dated 30th December 2007.
Let us modify the above problem to make things quantitative:
In the adjoining figure, SS is a spherical surface separating a transparent medium of refractive index 4/3 from air. C is the centre of curvature of the spherical surface. An observer located in air views the refracted image of an object AB of height 10 cm placed as shown in the medium of refractive index 4/3. If the distance of the object (from the spherical surface) is 3m, calculate the size of the image. We have n2/si n1/so = (n2 n1) /R.
Here n1 = 4/3, n2 = 1 (air), so = 3m, R = 2m.
Therefore, 1/si (4/3)/(–3) = [1– (4/3)] /2
The sign of the object distance so is negative since we have to measure if from P to A, opposite to the direction of the incident ray. The sign of R is positive since we have to measure it from P to C in the same direction as that of the incident ray.
The above equation gives si = 18/11 m.
The magnification is M = si/so = (–18/11)/ (– 3) = 18/33 = 0.545
Since the magnification is positive, the image is upright and virtual. We can obtain this information the moment we find that the sign of the image distance (si) is negative. [However, in the case of an image formed by reflection, the image distance will be negative for real and inverted image].
The size of the image = M× size of object = 0.545×10 cm = 5.45 cm.
(3) A straight rod of length very large compared to the focal length ‘f’ of a concave mirror lies along the principal axis of the mirror. The nearer end of the rod is at a distance ‘x’ from the pole of the mirror. If x > f, the length of the image of the rod is
(a) xf/(x f)
(b) xf/(x+ f)
(c) f2/(x+f)
(d) f2/x
(e) f2/(xf)
For a spherical mirror we have 1/f = 1/so + 1/si.
One end of the rod is far away from the mirror and hence the image of that end must be formed at the focus . [From the above equation, on applying the negative sign for f, you will get si = –f, the negative sign indicating that the image of the end is real and on the object side of the mirror].
The image of the near end of the rod will be formed at si given by
1/f = 1/x + 1/si so that
1/si = 1/f 1/x from which si = xf/(x –f)
[If you apply the negative signs for f and x, in accordance with the Cartesian sign convention, you will get si = xf/(x –f), the negative sign indicating that the image of the near end of the rod also is real and on the object side of the mirror].
The length of the image is [xf/(x –f)] – f = f2/(xf)
We will discuss more questions on geometric optics in the next post.