S5 Physics Paper: Optics, Lens Images And Refraction (100 Marks)

Physics – Optics and Lenses

S5 Physics: 6 questions, 24 parts, 100 marks, 75 minutes. Thin lens equation, refraction, telescopes, double slit interference and vision correction.

Question 1 Thin Lens Equation and Magnification (16 marks)

Three thin lenses with the focal lengths and object distances below are used on the same principal axis.

#	Question	Answer and marking notes
1	A convex lens of focal length 15.0 cm has an object at 45.0 cm. Find the image distance and magnification, and state whether the image is real or virtual.	1/v = 1/15.0 – 1/45.0 = 0.04444, v = 22.5 cm; m = -v/u = -0.50; a positive v gives a real, inverted, diminished image. (4 marks)
2	A convex lens of focal length 10.0 cm has an object at 15.0 cm. Find the image distance, magnification and the nature of the image.	1/v = 1/10.0 – 1/15.0 = 0.03333, v = 30.0 cm; m = -2.0, a real, inverted and magnified image. (4 marks)
3	A convex lens of focal length 12.0 cm has an object at 6.0 cm. Find the image distance and magnification, and explain the meaning of the signs.	1/v = 1/12.0 – 1/6.0 = -0.08333, v = -12.0 cm; m = +2.0; a negative v means a virtual image on the same side and a positive m means upright and magnified. (4 marks)
4	The first lens has a focal length of 15.0 cm. Find its power in dioptres and state its converging effect on parallel light.	P = 1/f = 1/0.150 = 6.67 D; a positive power means parallel light converges at the principal focus. (4 marks)

A beam travels from air into a glass block of refractive index 1.50; another beam travels inside the glass towards the glass-air boundary.

#	Question	Answer and marking notes
5	The angle of incidence is 30.0 degrees. Find the angle of refraction inside the glass (1 d.p.).	1.00 x sin 30.0 = 1.50 x sin r, sin r = 0.3333, r = 19.5 degrees. (4 marks)
6	Find the critical angle for light travelling from the glass to air (1 d.p.).	sin c = 1/1.50 = 0.6667, c = 41.8 degrees. (4 marks)
7	Find the speed of light in the glass (take the speed in vacuum as 3.0 x 10^8 m/s).	v = c/n = 2.0 x 10^8 m/s. (4 marks)
8	The wavelength in air is 600 nm. Find the wavelength in the glass and state whether the frequency changes.	Wavelength = 600 / 1.50 = 400 nm; frequency is set by the source and does not change at the boundary. (4 marks)
9	Diamond has a refractive index of 2.42. Find its critical angle and explain why a faceted cut makes it sparkle.	sin c = 1/2.42 = 0.4132, c = 24.4 degrees; the small critical angle gives repeated total internal reflection before the light leaves the top facet. (4 marks)

A refracting telescope has an objective of focal length 100.0 cm and an eyepiece of focal length 2.5 cm for observing distant objects.

#	Question	Answer and marking notes
10	Find the angular magnification of the telescope.	M = fo/fe = 100.0 / 2.5 = 40. (4 marks)
11	What is the separation of the two lenses when focused for a relaxed eye?	L = fo + fe = 102.5 cm. (4 marks)
12	If an eyepiece of focal length 2.0 cm is used, find the new angular magnification and tube length, and state the change in magnification.	M = 100.0 / 2.0 = 50 and L = 102.0 cm; magnification rises from 40 to 50, an increase of 25%. (5 marks)
13	Explain the advantage of a long objective focal length with a short eyepiece focal length, and give one main limitation of this type of telescope.	A long fo with a short fe gives a large fo/fe, raising magnification and light gathering; limitations include the weight of large lenses, chromatic aberration and a practical limit on aperture. (5 marks)

Monochromatic light of wavelength 600 nm falls on slits 0.50 mm apart; the screen is 2.0 m away.

#	Question	Answer and marking notes
14	Find the spacing between adjacent bright fringes in mm.	dx = wavelength x D / d = 2.4 x 10^-3 m = 2.4 mm. (4 marks)
15	How far is the fourth bright fringe from the centre?	The fourth fringe is four spacings away: 4 x 2.4 = 9.6 mm. (4 marks)
16	If light of wavelength 400 nm is used, find the new fringe spacing and state how the pattern changes.	dx is proportional to wavelength: new spacing = 2.4 x 400/600 = 1.6 mm, so the fringes become closer together. (4 marks)
17	Explain how this experiment supports the wave theory of light and what a single slit alone would show.	Alternating bright and dark fringes come from path difference and superposition of two coherent beams, a wave signature; a single slit gives one broad bright central diffraction band without equally spaced interference fringes. (4 marks)

Students measure four object and image distances u and v in cm with the same convex lens: (30.0, 30.0), (45.0, 22.5), (20.0, 60.0), (50.0, 21.0).

#	Question	Answer and marking notes
18	For each set, calculate 1/u + 1/v and the corresponding focal length f (1 d.p.).	The four values of 1/u + 1/v are 0.06667, 0.06667, 0.06667 and 0.06762; the focal lengths are 15.0, 15.0, 15.0 and 14.8 cm. (4 marks)
19	Find the mean of the four focal lengths (1 d.p.).	Mean = 59.8 / 4 = 14.9 cm. (4 marks)
20	If the lens is marked 15.0 cm, find the percentage error of the result (1 d.p.).	Percentage error = \|14.9 – 15.0\| / 15.0 x 100% = 0.7%. (4 marks)
21	Suggest two sources of error and how each can be reduced.	Uncertain image position (repeat readings and average) and misalignment of the ruler or lens centre with the axis (align the optical axis first). (4 marks)

Two students have these data: student A has a far point of 0.50 m and student B a near point of 1.00 m.

#	Question	Answer and marking notes
22	Student A is short sighted. Find the power of the correcting lens and state the lens type.	Light from infinity must appear to come from 0.50 m: P = -1/0.50 = -2.0 D; a negative power means a diverging lens. (5 marks)
23	Student B is long sighted and wants to read at 0.25 m. Find the power of the correcting lens.	P = 1/0.25 – 1/1.00 = 4.0 – 1.0 = +3.0 D. (5 marks)
24	Compare the causes of short sight and long sight and state where the image forms on the retina in each case.	Short sight usually comes from an elongated eyeball or a lens that is too strong, so parallel rays focus in front of the retina; long sight comes from a short eyeball or a weak lens, so rays focus behind the retina. (4 marks)

A TutorZone practice paper; original questions and passages, not publisher material.