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## Friday, August 9, 2013

### AP Physics B - Multiple Choice Practice Questions on Geometric Optics

– 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
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