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What should be the distance between the object in Exercise 30 and the magnifying glass if the virtual image of each square in the figure is to have an area of "6.25 mm"^(2). Would you be able to see the squares distinctly with your eyes very close to the magnifier ? [Note : Exercises 29 to 31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.' |
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Answer» Solution :Here, MAGNIFICATION in area = 6.25 linear magnification `m=sqrt(6.25)=2.5` As `m=(upsilon)/(u) or v=m u=2.5u` As `(1)/(upsilon)-(1)/(u)=(1)/(f)` `(1)/(2.5u)-(1)/(u)=(1)/(10)` `(1-2.5)/(2.5u)=(1)/(10)` `u=-6cm` `y=2.5u=2.5(-6)=-15CM` as the virtual image is at 15 cm, where as distance of DISTINCT vision is 25CM, therefore, the image cannot be SEEN distinctly by the eye. |
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