Question

What should be the distance between the object in Exercise 9.30 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 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Answer 1

now we want the area of square shaped virtual image as 6.25 mm² .

so, side length of image is I = $\sqrt{6.25}$ = 2.5mm .

for the given magnifying lens of focal length 10cm , we can calculate the required position of object.

so, magnification, m = I/O = v/u

or, I/O = v/u

or, (2.5mm)/(1mm) = v/u , so v = 2.5u [ see exercise 9.29 , here O = 1mm ]

from lens formula,

1/v - 1/u = 1/f

or, 1/2.5u - 1/u = 1/+10

or, 2/5u - 5/5u = 1/10

or, -3/5u = 1/10

or, u = -30/5 = -6cm

so, v = 2.5(-6) = -15cm

thus, the required Virtual image is closer than normal near point .

thus, the required Virtual image is closer than normal near point .thus the eye cannot observe the image distinctly.

also read

(a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly wi...

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