An object is placed at x = 0. A converging lens with f1 = 35.0 cm is placed at x
ID: 1412485 • Letter: A
Question
An object is placed at x = 0. A converging lens with f1 = 35.0 cm is placed at x = 60.0 cm and a concave mirror with radius of curvature R = 48.0 cm is placed at x = 120 cm. Considering light that leaves the object, passes through the lens, reflects from the mirror and then passes back through the lens, determine:
(a) the x-coordinate of the final image;
(b) the overall magnification of the final image;
(c) whether the final image is real or virtual;
(d) whether the final image is inverted or upright.
A drawing would be very helpful :) ANSWERS:-69.2 cm, 1.88, real, upright
Explanation / Answer
the object distance u=60 cm
focal length of lens 1=>f1=35 cm
focal length of lens 2=>f2=24 cm
the distance b/w two lenses d=120-60=60 cm
this concept belongs to rayoptices in light
first we find the image formed b/w two lenses
1/v1-1/u=1/f1
1/v1-1/60=1/35
1/v1=1/60+1/35
v1=35*60/60+35 =2100/95=22.1 cm
the image formed b/w the lenses=v1=22.9 cm
now the object distance for second lenses u2=d-v1=60-22.9=37.1 cm
now we find the image formed in second lenses
1/v2=1/u2-1/f2=1/37,8-1/24.
v2=912/13.1.=-69.4 cm
the finial image is formed =-69.4 cm
(b) ans
the over all magnifaction of lens >m=69.4/37.1=1.87
(c) ans
the finial image formed is real because the magnifaction is positive
(d) ans
the finial image formed is upright because the magnifaction is negative