an object of height 4cm is paced in front of a diverging lens of focal length 40 cm if the object distance 1360 cm find the position,height,magnification of image?
Answers
Answer:
focal length = f = - 15 cm
object distance = u = -25 cm
image distance = v = ?
hi = ?
ho = 4 cm
Mirror formula,
1/v + 1/u = 1/f
1 over f space equals space 1 over u plus 1 over v
rightwards double arrow fraction numerator 1 over denominator negative 15 end fraction equals fraction numerator 1 over denominator negative 25 end fraction plus 1 over v
rightwards double arrow space fraction numerator 1 over denominator negative 15 end fraction plus 1 over 25 space equals space 1 over v
rightwards double arrow space 1 over v equals fraction numerator 25 minus 15 over denominator negative 375 end fraction
rightwards double arrow space v space equals negative space 37.5 space c m space
m equals fraction numerator negative v over denominator u end fraction equals h subscript i over h subscript o
rightwards double arrow space m space equals fraction numerator negative left parenthesis negative 37.5 right parenthesis over denominator negative 25 end fraction equals h subscript i over 4
h subscript i space equals space 150 over 25 equals space minus 6 space c m
Thus, to get image of object sharp it has to be placed at 37.5 cm
Size of image is 6 cm
As the height of object is more and negative, nature of image is real, inverted and magnified.
Explanation:
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h = 4 cm u = -15 cm P = -10 D
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