Figure 27a3See Answer T5. A convex lens used as a magnifying glass.

Transcription

1 F1 Figure 27a (in Answer T5) shows a diagram similar to that required, but with different dimensions. The object is between the first focus and the lens. The image is erect and virtual. The lateral magnification m = v/u : m = ( 251cm)/( 101cm) = (v is the image distance, u is the object distance and the Cartesian sign convention has been used.) (a) cm F 2 Figure 27a3See Answer T5. A convex lens used as a magnifying glass.

2 F2 Use the thin lens equation (Equation 12): The thin lens equation (Cartesian sign convention) 1 v 1 u = 1 f (Eqn 12) 1/v 1/u = 1/f, with u = 251cm, f = +101cm, so that: 1/v = (1/101cm) + ( 1/251cm) = (5 2)/(501cm) = 3/(501cm) and v = +501cm/3 = cm. The image is real and inverted with a magnification m = v/u = 16.71cm/251cm = 0.67.

3 B X 0 5 cm object O h C F 2 F 1 I h' image u Y v M Figure 193Ray diagram for a convex lens with f = 101cm and u = 151cm. Note that the zero on the scale bar does not correspond to the origin of Cartesian coordinates which is at C. Your sketch should resemble Figure 19 but with the object further from the lens, and the image closer and smaller than the object.

4 F3 The ray diagram is similar to Figure 24. Use the formula 1/u + 1/v = 1/f with u = 151cm and f = 101cm. B (1) (2) β β P h Then 1/v = (1/ 101cm) (1/ 151cm) = ( 3 + 2)/(301cm)= 1/(301cm), and v = 301cm. The image is real, inverted and 301cm in front of the mirror. I R (1) (2) β O 2β F α α C M f u r v Figure 243A spherical concave mirror forming a real inverted image of an extended object.

5 Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to Ready to study? in Subsection 1.3. Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Closing items. If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it here.

6 R1 (i) The incident ray, the reflected ray and the normal at the reflection point all lie in the same plane. (ii) The angle of the incident ray to the normal equals the angle of the reflected ray to the normal.

7 R2 Snell s law says: µ air 1sin1θ air = µ glass 1sin1θ glass. The ray is refracted towards the normal since µ glass > µ air We have: θ glass = arcsin sin θ air µ glass = arcsin θ glass = 9.9 ( ) = arcsin ( ) ( ) The refractive index µ glass = c/v, where v is the speed of light in the material and c is the speed of light in a vacuum. v = m1s 1 /1.5 = m1s 1

8 R3 At 10, tan1θ = and θ = 10π/180 = radians. The percentage error is: (0.0018/0.1763) 100% = 1%. At 10, sinθ = and the percentage error is: (0.0009/0.1736) 100% = 0.5%. The approximation improves as θ gets smaller.

9 T1 The apex angle A of the prism is 60. Using Equation 5 for D min : D min = 2 arcsin0[µ sin(a/2)] A (Eqn 5) blue light: d min = = green light: d min = = yellow light: 0000D min = = red light: d min = = 37.35

10 T2 In the paraxial approximation, µθ = µ θ From the triangles of Figure 10 we have θ = β α and θ0 = β γ, so that µ(β α) = µ (β γ) Replacing angles by their tangents then gives β = tan1β = h/r, α = tan1α = h/l, γ = tan1γ = h/l so µ h r h l and µ l + µ l = µ h r h l = µ µ r This differs from Equation 7 µ µ µ µ + = l l r in the sign of the first term. (Eqn 7)

11 θ' µ θ µ' O α γ β O' R C l l' r Figure 103The point image of a point object formed by a concave boundary surface (see Question T2). T3 Referring to Figure 10 we see that O, O and R all lie to the left of the origin at C so that l, l and r must all be made negative. With these changes, the equation which forms the answer to Question T2 is unchanged.

12 T4 Use the lens maker s equation: 1/f = (µ 1)(1/r 1 1/r 2 ) (Eqn 11) Lens A: 1/f = 0.5[(1/25) (1/15)]1cm 1 = [0.5(3 5)/75]1cm 1 = ( 1/75)1cm 1. So f = 751cm. The lens is diverging with convex front face and concave back face. (See Figure 26a.) Lens B: 1/f = 0.5[(1/ 10) (1/ 5)]1cm 1 = [0.5( 1 + 2)/10]1cm 1 = (1/20)1cm 1. So f = 201cm. The lens is converging with concave front face and convex back face. (See Figure 26b.) Lens C: 1/f = 0.5[(1/± ) (1/ 20)]1cm 1 = 0.5(0 + 1/20)1cm 1 = (1/40)1cm 1. So f = 401cm. The lens is planoconvex and is converging. Note that the radius of curvature of a plane surface may be taken as ± ; it makes no difference. (See Figure 26c.) Lens D: 1/f = 0.5[(1/20) (1/20)]1cm 1 = 01cm 1. So f =. The front surface is convex and the rear surface is concave with equal curvature. There is zero net refraction and rays pass through with no change of direction. (See Figure 26d.) (a) (c) (b) (d) Figure 263See Answer T4.

13 T5 (a) This is an example of a convex lens used as a magnifying glass. Using Equation 12 1 The thin lens equation v 1 u = 1 (Eqn 12) f with f = 101cm and u = 61cm gives 1/v = (1/101cm) + (1/ 61cm) = (3 5)/(301cm) = 2/(301cm) so that v = 151cm. The ray diagram is shown in Figure 27a. (a) cm F 2 Figure 27a3See Answer T5. A convex lens

14 (b) This is more difficult to visualize, but Equation 12 1 The thin lens equation v 1 u = 1 (Eqn 12) f takes care of the calculation without any difficulties. As the object is virtual, it must be to the right of the lens, so that u = +161cm. The focal length f is +201cm and the equation gives 1/v = (1/201cm) + (1/161cm) = (4 + 5)/(801cm) = 9/(801cm), so that v = 8.91cm. P Q (b) X C M I real image B O virtual object cm Figure 27b3See Answer T5. A convex lens produces a real image of a virtual object. F 2 To construct the diagram we must choose two rays which would have met at the tip B of the arrow if the lens were not there. Suitable rays are PX and QC in Figure 27b. QC proceeds undeviated to B but PX, being parallel to the optical axis is refracted through F 2. The image of the arrow tip is formed at the intersection M of XF 2 and CB.

15 T6 It is a test of the care you have taken with your ray diagrams to see how closely your measured results come to the calculated ones. The precise results obtained from calculation are as follows: For Figure 19, u = 151cm, v = 301cm, m = 301cm/( 151cm) = 2, and the minus indicates an inverted image For Figure 20, u = 151cm, v = 61cm, m = 61cm/( 151cm) = 0.4 and the image is erect. For Figure 21, u = 801cm, v = 1331cm, m = 1331cm/801cm = 1.66 with an inverted image.

16 T7 Refer to Figure 28. It is necessary to draw two sets of principal rays. The first set is used to locate the first image and the second set to locate the final image produced by the second lens with the first image as the second object. M 1 I 1 O B I 2 M 2 10 cm Figure 283See Answer T7.

17 For the first lens: 1/v 1 = 1/f 1 + 1/u 1 = (1/101cm) + (1/ 51cm )= 1/(101cm) so v 1 = 101cm The first image is virtual and erect. The distance of the second object (i.e. the first image) to the second lens is: u 2 = (d v 1 ) = [35 ( 10)]1cm = 451cm 1/v 2 = 1/f 2 + 1/u 2 = (1/101cm) + (1/ 0451cm) 11 1= (9 2)/(901cm) = 7/(901cm) so v 2 = 12.91cm The final image is real, inverted and to the right of the second lens. The overall magnification is given by the product of the separate magnifications of the two lenses: m = m 1 m 2 = (v 1 /u 1 )(v 2 /u 2 ) = [( 101cm)/( 51cm)][12.91cm/( 451cm)] = The final real image is reduced in size by a factor 0.57 and it is inverted as the minus sign indicates.

18 T8 Use Equation 19: 1 u 1 v = 1 f (Eqn 19) with f = 201cm and v = 151cm. Then 1/u = ( 1/201cm) + (1/151cm) = ( 3 + 4)/(601cm) = 1/(601cm), and u = 601cm. Remember, this result made no use of the Cartesian sign convention.

19 T9 Use Equation 21a: 1 v + 1 u = 1 f (Eqn 21a) with f = 201cm. (a) u = 301cm so 1/v = (1/ 201cm) (1/ 301cm) = 1/(601cm) and v = 601cm The image is in front of the mirror and therefore is real. (b) Now we have u = 151cm and the same focal length so 1/v = 1/( 201cm) 1/( 151cm) = 1/(601cm) to give v = +601cm so that the image is to the right of the mirror and is therefore virtual. (c) With u = 0, 1/v = 1/( 201cm) 1/( 0 ) = 1/(201cm) and v = 201cm. The image is real and at the focus.