1031_Physics ProblemsTechnical Physics

1031_Physics ProblemsTechnical Physics - 372 P36.62 Image...

This preview shows page 1. Sign up to view the full content.

372 Image Formation P36.62 111 1 10 0 1 12 5 qf p =−= .. cm cm so q 1 50 0 = . cm (to left of mirror). This serves as an object for the lens (a virtual object), so 1 16 7 1 25 0 222 p cm cm af and q 2 50 3 =− . cm , meaning 50.3 cm to the right of the lens. Thus, the final image is located 25 3 . cm to right of mirror . M q p M q p MMM 1 1 1 2 2 2 12 50 0 400 50 3 25 0 201 805 == . . . . . . cm 12.5 cm cm cm a f a f Thus, the final image is virtual, upright , 8.05 times the size of object, and 25.3 cm to right of the mirror. P36.63 We first find the focal length of the mirror. 1 10 0 1 800 9 40 0 fpq =+= + = . cm cm cm and f = 444 c m . Hence, if p = 20 0 cm, 1 1 20 0 15 56 88 8 qfp = . . cm cm cm . Thus, q = 571 c m , real. *P36.64 A telescope with an eyepiece decreases the diameter of a beam of parallel rays. When light is sent through the same device in the opposite direction, the beam expands. Send
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online