9.25 What should be the distance between the object in Exercise 9.24 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

[Note: Exercises 9.23 to 9.25 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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8 months ago

9.25 Area of the virtual image of each square, A = 6.25 mm2

Area of each square,  A0 = 1 mm2

Hence the linear magnification of the object can be calculated as:

m = AA0 = 6.251 = 2.5, but m = vu or v = 2.5u

Focal length of the magnifying glass, f = 10 cm. According to lens formula

1f = 1v - 1u

110 = 12.5u - 1u or 110=-1.52.5u . Hence u = -6 cm and v =

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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