P3659 A hemisphere is too thick to be described as a thin lens The light is

P3659 a hemisphere is too thick to be described as a

This preview shows page 345 - 348 out of 618 pages.

P36.59 A hemisphere is too thick to be described as a thin lens. The light is undeviated on entry into the ﬂat face. We next consider the light’s exit from the second surface, for which R = − 6 00 . cm . The incident rays are parallel, so p = ∞ Then, n p n q n n R 1 2 2 1 + = becomes 0 1 1 00 1 56 6 00 + = q . . . cm and q = 10 7 . cm P36.60 (a) I r = = × ( ) = P 4 4 50 4 1 60 10 1 40 2 2 2 π π . . . W m kW m 2 (b) I r = = ( ) = P 4 4 50 6 91 2 2 π π . . W 4 7.20 m mW m 2 (c) 1 1 1 p q f + = : 1 7 20 1 1 0 350 . . m m + = q so q = 0 368 . m and M h q p = = − = − 3 20 0 368 . . cm m 7.20 m ′ = h 0 164 . cm (d) The lens intercepts power given by P = = × ( ) ( ) IA 6 91 10 4 0 150 3 2 . . W m m 2 π and puts it all onto the image where I A = = × ( ) ( ) P 6 91 10 15 0 4 0 164 3 2 . . . W m cm 2 π π cm ( ) 2 4 I = 58 1 . W m 2 FIG. P36.59
346 Chapter 36 P36.61 From the thin lens equation, q f p p f 1 1 1 1 1 6 00 12 0 12 0 6 = = ( ) ( ) − − . . . . cm cm cm 00 4 00 cm cm ( ) = − . When we require that q 2 → ∞ , the thin lens equation becomes p f 2 2 = . In this case, p d 2 4 00 = − − ( ) . cm Therefore, d f + = = 4 00 12 0 2 . . cm cm and d = 8 00 . cm P36.62 (a) For the light the mirror intercepts, P = = I A I R a 0 0 2 π 350 1 000 2 W W m 2 = ( ) π R a and R a = 0 334 . m or larger (b) In 1 1 1 2 p q f R + = = we have p → ∞ so q R = 2 M h h q p = = − so ′ = − = − ° ° h q h p R 2 0 533 . π rad 180 = − ( ) R 2 9 30 . m rad where h p is the angle the Sun subtends. The intensity at the image is then I h I R h I R R a a = = = ( ) × P π π π 2 0 2 2 0 2 2 3 4 4 4 2 9 30 10 . r ad ( ) 2 120 10 16 1 000 9 30 10 3 2 2 3 × = ( ) × ( W m W m rad 2 2 R R a . ) 2 so R R a = 0 025 5 . or larger FIG. P36.61
Image Formation 347 P36.63 For the mirror, f R = = + 2 1 50 . m . In addition, because the distance to the Sun is so much larger than any other distances, we can take p = ∞ . The mirror equation, 1 1 1 p q f + = , then gives q f = = 1 50 . m Now, in M q p h h = − = the magnification is nearly zero, but we can be more precise: h p is the angular diameter of the object. Thus, the image diameter is ′ = − = ( ) ° ( ) = − h hq p 0 533 1 50 0 0 . . . o π rad 180 m 14 0 1 40 m cm = . P36.64 (a) The lens makers’ equation, 1 1 1 1 1 2 f n R R = ( ) + becomes: 1 5 00 1 1 9 00 1 11 0 . . . cm cm cm = ( ) ( ) n giving n = 1 99 . . (b) As the light passes through the lens for the first time, the thin lens equation 1 1 1 1 1 p q f + = becomes: 1 8 00 1 1 5 00 1 . . cm cm + = q or q 1 13 3 = . cm , and M q p 1 1 1 13 3 1 67 = − = − = − . . cm 8.00 cm This image becomes the object for the concave mirror with: p q m = = = 20 0 20 0 13 3 6 67 1 . . . . cm cm cm cm and f R = = + 2 4 00 . cm The mirror equation becomes: 1 6 67 1 1 4 00 . . cm cm + = q m giving q m = 10 0 . cm and M q p m m 2 10 0 1 50 = − = − = − . . cm 6.67 cm The image formed by the mirror serves as a real object for the lens on the second pass of the light through the lens with: p q m 3 20 0 10 0 = = + . . cm cm The thin lens equation yields: 1 10 0 1 1 5 00 3 . . cm cm + = q or q 3 10 0 = . cm and M q p 3 3 3 10 0 1 00 = − = − = − .

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