Question

A system of two lenses forms an image of an arrow at x = x3 = 57.9 cm. The first lens is a diversing lens located at x = 0 and has a focal length of magnitude f. = 12.4 cm. The second tens is located atx=27=25.8 cm and has an unknown focal length. The tip of the object arrow is located at (x,y) = (x, y) = (-36.6 cm, 12.1 cm). 1) What is xi, the x co-ordinate of image of the arrow formed by the first lens cm Submit 2) What is y, they co-ordinate of the image of the tip of the arrow formed by the first lera? cm Submit 3) What is f2, the focal length of the second lens. If the lens is a converging lens, f is positive. If the lens is a diverging lens, 2 is negative. cm Submit 4) What is y), the y co-ordinate of the image of the tip of the arrow formed by the Dno lens system cm Submit 5) The positions of the two lenses are now interchnaged fi.e., the second lers is moved to x = 0 and the diverging lens is moved to xx2.25.8 cm). What is the nature of the final image in this new System Real and Inverted Real and Upright Virtual and Inverted Virtual and Upright Submit

Answer 1

m = hi= mha=0.3179 X 1201 - 3.847cm ist lens 2nd hreal image X = 57.9cm fi= -12.4cm x2 =25.8cm 154 lens U = -36.6 cm f = -12.4cm - - Vpt $$+ke t =) u=f4 ftu g) V- -12.4% -36-6 = - 110 6369 cm - 36.6 -12.4 m= - 11.6369 >15 = 0.3179 - 36.6 b) hi ho Y = 3.847cm e) 2nd lens U= - (25.84 11.6369) = - 37.44cm V=3201cm -te t-1/43 fa ur - 370466 32.1 - 17.28cm - 37.44-3201 u-V m₂ = u 32.1 - 37.44 - - 0.857 Y = may = – 0.857 X 8.847 = -3+298 cm Answer

Answer 2

m = hi= mha=0.3179 X 1201 - 3.847cm ist lens 2nd hreal image X = 57.9cm fi= -12.4cm x2 =25.8cm 154 lens U = -36.6 cm f = -12.4cm - - Vpt $$+ke t =) u=f4 ftu g) V- -12.4% -36-6 = - 110 6369 cm - 36.6 -12.4 m= - 11.6369 >15 = 0.3179 - 36.6 b) hi ho Y = 3.847cm e) 2nd lens U= - (25.84 11.6369) = - 37.44cm V=3201cm -te t-1/43 fa ur - 370466 32.1 - 17.28cm - 37.44-3201 u-V m₂ = u 32.1 - 37.44 - - 0.857 Y = may = – 0.857 X 8.847 = -3+298 cm Answer