Final Exam of 8/5/2010

(last exam in sample exams book)

SOLUTIONS

1. M =hi/ho -q/p = 2/0.3 ==> p= 3 (0.3)/2 = 0.45m

1/f = 1/p + 1/q = 2.22 + 0.333 = 2.55 ==> f = 0.39 m

2. ml = d sin q; d = W/N . m = 2 for 2nd order, so N = (W sin q)/2l

3. m = -q/p = 5 ==> q = 5p = 5x40 = 200; 1/40 + 1/200 =1/f ==> f= 200/6

for 2X, q = 2p ==> 1/p + 1/2p = 6/200 ==> p = 50 cm

Initially p = 40 cm, so move it 10 cm farther away.

4. To see far-away objects, 0 - 1/2m = 1/f ==> f = -2m = -200 cm

With this lens, we want to know where the object will be when q= original near point, so 1/p - 1/20 = -1/200 ==> 1/p = (10 - 1)/200 ==> p = 22cm

5. The silvered sphere is a diverging mirror with R = -15cm, so

1/2 + 1/q = -2/0.15 -1/2 ==> 1/q = -13.8 ==> q = 0.0723

m = -q/p = 0.0723/2 = 0.039

6. See § 22.5

7. See § 25.6

8. Observing very distant (thus, low intensity) objects requires maximum light collection, in order for the image to be bright enough to be visible.

Large diameter objective = more photons intercepted = brighter image

9. Light of intensity I passes through A and emerges with intensity I cos² q_A .

it then passes through B, emerging with intensity I cos² q_A cos² q_B

Thus 0.15 I = I cos² q_A cos² q_B . But cos² q_A = (cos 40º)² = (0.766)² = 0.586

So 0.15 = cos² q_A (0.586) ==> cos² q_A = 0.15/0.586 = 0.256 ==> cos q_A = 0.5056; thus q_A = 59.6º or 60º

10. There is a phase shift at both the upper & lower surface, so

2nt = (m + ½ ) l . m=0 for thinnest film ==> t = 0.5 (550 nm)/(2 x 1.33)

= 103 nm

11. f = 40mm = 0.04m; thus 1/0.54 + 1/q = 1/0.04 ==> 1/q = 25 - 1.85 = 23.15

==> q = 0.0432. The difference in mm is 43.2 - 40 = 3.2 mm increase

12. ml = 2d sin q . Spreading out the fringes requires an increase in q.

sin q = ml/2d so either increase l or decrease d. Of the options, only (1) works.

13. Diffraction of waves requires edges or apertures with dimensions roughly comparable to wavelength. Radio waves have l 's of the order of many meters,

thus buildings can diffract them.

14. The ray from the viewed location must emerge from the water surface at slightly less than the critical angle. So the critical angle is the limit, where the ray to the observer travels along the water surface.

sin q_C for water = 1/1.33 = 0.752 ==> q_C = 48.75º

The ray up from the spot that can be seen is the hypotenuse of a triangle with y = depth and x = 5 m. tan q_C = x/y ==> y = 5 m/1.14 = 4.38 or 4.4 m

15. The intensity I = power/area = 50 kW/(4pr²) ; r= radius of wave front = 150 km

I = 5 e4/[4p x (150 e3)² = 0.177 or 0.18mW/m²