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A 4.0 cm tall object is placed 50.0 cm from a diverging lens having a focal...

A 4.0 cm tall object is placed 50.0 cm from a diverging lens having a focal length of magnitude 20.0cm. a. What is the location of the image? b. What is the magnification of the image? c. Is the image inverted or upright? d. Is the image virtual or real?

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Answer #1

Sol:

1/d0+1/di= 1/f
=> 1/du= 1/f - 1/d0
= -1/20cm -1/50cm
=-0.005-0.02=-0.07 cm
=> di = -14.285cm
Minus sign means image is on same side as object, and therefore it must be virtual.
m = -di/d0
= -(-14.285)/50
= 0.2857
So, +0.2857,m=h0/hi
=> h0= m*hi
= 0.2857*4cm
= 1.143 cm
+ means erect, so image is upright and 1.143cm tall.
a) location: same side as object
b) magnification : +0.2857
c) image is UPRIGHT
d) image is VIRTUAL

Hope will get VOTED
thank you

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