Question

Part A In the figure (Figure 1), what is the orientation of the final image? virtual, upright virtual, inverted real, upright real, inverted O O Figure < 1 of 1 Submit Request Answer Part B fi = 10 cm f2 = -30 cm What is the position of the final image? 1.0 cm O to the right of the lens O to the left of the lens 5.0 cm 5.0 cm lens mirror Submit Request Answer

Answer 1

solution J = 10cm I2 = -30cm 1.0cm T mirror 5.0cm Toom = – 5.0 am Here object distance from lens using lens formula, - I - 25cm - Tom = t - Ś I = 1 - 2 = 1 16 Here object Now we from object = - 10cm shows image is formed the lenso. That will be distance from mirror same side of upright & virtual. U₂ - - 10cm+ (-50m) ₂ - - 15cm wsing mirror formula I + 1) = -30 -+ is - 1 +2 Scanned with CamScanner 30

& positine value of position of image (Jinal) represents that image is formed another side of mirror. Therefore its nature Corientation) is. virtual & upright. option ② is correct, of o As position final image is 30cm to the right of mirror. Therefore it will obviously, be to the right of the lens. option ® is correct. ② As we got above, position of final image from mirror - +30cm . The distance between the final image & the lens = 5 cm + 30 cm – 135 cm Ang As we have , magnification of minor m = hi Cheight of image) - - - to (height of object) height of image (h) = - Cho u = - 30 x 100 Tu = 3.0fcm Ang or Scanned with CamScanner
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