Question

3. (0/10 Points) DETAILS PREVIOUS ANSWERS TAMUCOLPHYSEML1 5.POST.003. MY NOTES A screen is placed a distance d = 39.0 cm to the right of a small object. At what two distances to the right of the object can a converging lens of focal length +7.85 cm be placed if the real image of the object is at the location of the screen? (Hint: Use the thin lens equation in the form si, SO 51 - 9. - st. Write s'=d-s and solve fors in the resulting quadratic equation.) shorter distance x cm longer distance cm For this lens, what is the smallest value d can have and an image be formed on the screen? cm Additional Materials

Answer 1

solo of 78 cm -d: 39 cm 4 39-x- x Screen v- uf cetf t = 12 + 1 2 + = (39-x) = utf uf - 1x7.85 -x+7.85 (89) [-x +7-25) = -4-85 * +22 -392 +39x7.85 - 7.852= - 7.854 → 22 x² – 39x + 306.15-0. n +39 ± √ 39² W 2x1 - +39 + 17.22 28.11 ; 10.89 cm $ 4x306.15 89€ re Shorter distance = 10.89 cm longer distance = 28.11cm for real image & urf image object must be at a distance minimum value of da 24 - 2x7.85 c 15.70cm. d=15.70 cm