Practice Problem 24.10 s equation to a diverging lens, and then we will confirm

Question

Practice Problem 24.10 s equation to a diverging lens, and then we will confirm our results by constructing the image using a you are given a thin diverging lens and you find that a beam of parallel rays spreads out after passing ens, as though all the rays came from a point 20 0 cm from the center of the lens. You want to use this lens to fornm one-third the height of a real object. (a) Where should the object be placed? (b) Draw a principal- an upright, virtual image that is ray diagram. SOLUTION SET UP AND SOLVE Part (a): The behavior of the parallel incident rays indicates that the focal length is negative =-20.0 cm. We want the lateral magnification to be m = + 1/3 (positive because the image is to be upright) From the definition of magnification, m = 1/3 =8 '/s So we use s' =-3/3 n the thin lens equation + 8s/3 20.0 cm 40.0 cm s' 40.0 cm=-13.3 cm1 = Part (b): shows our principal-ray diagram for this problem. REFLECT In part (a), the image distance is negative, so the real object and the virtual image are on the same side of the lens -13.3 cmm K-200cm200cm- 40.0cm-

Explanation / Answer

Given that

focal length f=-13 cm

hight of the image hi=ho/2

now we find the magnifaction of the lens

magnifaction m=hi/ho=ho/2/ho=1/2

magnifaction m=1/2

now we find the distance of the object placed

magnifaction m=f-u/fu/u

1/2=f-u/fu^2

fu^2=2f-2u

-13u^2=2*-13-2u

13u^2-2u-26 =0

compare this equations ax^2+bx+c=0

u=2+[2^2+4*26*13]^1/2/2*13

=2+36.8/26

=1.5 cm

=>the object distance u=1.5 cm

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