Question

Consider the optical system shown in the figure. The lens and mirror are separated by d = 1.00 m and have focal lengths off, = +89.0 cm and f2 = -48.8 cm, respectively. An object is placed p = 1.00 m to the left of the lens as shown. Object Lens Mirror 1.00 m 1.00 m i (a) What is the distance (in cm) and location of the final image formed by light that has gone through the lens twice? (Give the magnitude of the distance and select its location with respect to the lens.) distance location cm to the right of the lens (b) Determine the overall magnification of the image. (c) State whether the image is upright or inverted. O upright O inverted

Answer 1

0.89 m im 9) using lens formula 1 tv ta 1 V -1 0.89 I V 0.89 7.09 m 8.09m Image formed by lens will act object for a = 8.09-1 mirror formula + 52 using U - 48.8 1 + 1 + 709. -1 2 48.8 709 va VO - 45.6 cm 5) Distance = 45.6 cm locatron to the left of mirror mager freation (m) mix m2 magnifreation produced lens ma - mirror my PA

=) 8.09 - 8.09 0.064 2) ma= -112 - (-48.6) 709 U. magnification (m) = – 8. 00 x 0.064 m m - - 0.52 means Final m Is negative it inverted 1 image