Question

Q1 Thin Film Engine oil (n = 1.50) has spilled onto a smooth flat lake (n = 1.33), spreading into a thin surface film of thickness, d. Sunlight illuminates the lake, reflecting from an incident angle of 45° to your eye. Your eye detects sunlight in the 400 to 740 nm wavelength range. (a) [3 mark] State an equation for wavelength, 2m, that satisfies the bright intensity fringe condition at the 45° viewing angle. (b) [2 marks] Determine the minimum film thickness below which interference maxima will not be observable for the 45° viewing angle. c) [2 marks] Determine all solar wavelengths that your eye will see strongly reflected at the 45° viewing angle when the film thickness is 0.8 um? (d) [2 marks] Use the spectral response in (c) to estimate the maximum film thickness beyond which interference fringes will not be observable from the oil film for the 45° viewing angle? (e) [1 mark] You are given a laser of 630 nm wavelength. What angle(s) with respect to the surface normal must you point the laser in order to induce bright fringe maxima for reflection by the 0.8 Um thick film? (f) [1 bonus mark] What polarization state of laser in (e) will produce the brightest reflection? sunlight eye Nair = 1.0 no = 1.5 Ź Tw = 1.33

Answer 1

Question ti. Engine oil (no=15) has spilled onto a smooth blat dake Cow = 1,33), spreading into a thin film surface bilm of thickness d. The sun light falls at on incident angle 45° to our ege. a) The condition for brightness I interference maxima O ano de 2m d = minimum thickness o ama wavelength of light- or, am = 4 x1 " I am Am * = 6 denil Camp domin b) For am = 400 nm , d = 400 nm a prom 66.67 nm 123.33mm For am - 740 nm , d= 740 nm - Tout c) when the thickness of the to da 0.8 um film is given by, 0,80100 m - so. a = 6 x 0,8%100 pm 16 - 4800 nm (Ann ... - B rom (An d) For maximum the ckness id = 1 . cos 450 manumum LSS 4M 4800 x 109 OUD or da x - m - 5 oo = 1.13 pa (Ans cum e) For a given laser of 630 nm wavelength the angles which makes with the surface normal is, 2 ne ol Cose - or, Coso = f = 6300109 . And 4x1.500.87160 Coso = 0.13 - Cos 82.46 L O - 82469 Camp