At the very bottom where they calculate Hi=Mho, why are 64.7 and 190 the same si
ID: 1491860 • Letter: A
Question
At the very bottom where they calculate Hi=Mho, why are 64.7 and 190 the same signs? One is the image distance of a converging mirror which would be negative, and the other is the image distance of a diverging lens, which should be positive.
41 .. A 2.05-cm-tall object is placed 30.0 cm to the left of a converging lens with a focal length fi= 20.cm. A diverging lens, with a focal length -42.5 cmis placed .0 cm to the f2 = 11, 30 right of the first lens. How tall is the final image of the object? A oritrea Ori move P1 N fi d, ol d1 an - (fi dol ) ( 20.5 cm 30.0 Cm ) d o2 –64.7 cm +30.0 cm =--34.7 cm 64.7 cm obsOr itroge of top time, dao 30th obey a Harry do in-- 42 27 on = 190 cm (f2 do2 ) (--42.5 cm --34.7 cm) 20 T cr d; i2 No _ (64.7 cm (190 cm) (2.05 cm) h = mh, = mm2 h . = =-24.2 cm T d ol o2 o1002 (30.0 cm)–34.7 cm) The final image is inverted and (24.2 cm tall. did (|Explanation / Answer
for a converging lens the image ditance is 64.7 cm is positive and for diverging lense already we have taken object distance as negative ,So we got final imge distance as positive
if we take the object distance for diverging lens as a positive ,we get the image of diverging lens as negative
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