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Unformatted text preview: Right-to-Left Propagation - Refraction Light is propagating from right to left (rather than the standard left to right).
An object is located 200 mm to the right of the spherical refractive surface as shown in
the ﬁgure. The object is in air (index of refraction of 1.0). To the left of the refractive
surface is glass of index of refraction of 1.5. The spherical refractive surface has a radius
of curvature of 50 mm. The center of curvature is located to the left of the surface. Spheﬁcal
Surface A Calculate, showing all work, the location and magniﬁcation of the image due to
_ the object shown. Express your answers accurately to four signiﬁcant ﬁgures. Specify
whether the image is left or right of the spherical surface. Specify whether it is a real or
virtual image. Put your ﬁnal answers in the spaces provided. Using a distance between tick marks of 50 mm, draw on the above diagram, the
parallel ray, the chief ray, the focal ray, and the image. Magnitude of image distance from spherical surface : mm
The image is to the (left) (right) of the spherical surface. (Circle one.)
Magnitude of linear magniﬁcation = The image is (real) (virtual). (Circle one.) Right-to-Left Propagation - Refraction Reverse left and right and proceed using standard methods R = +50mm, 77.1 = 1.00, 712 = 1.50, s 2 +200mm f1 = M— : +100mm
n2 — n1
f2 = _n2___ = +150mm
n2 — n1
n1 2 __ 712-711
3 s’ _ R '
m = — ms 2 ——1.00 (inverted real image) n23 ...
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- Spring '08